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When you initially learned integration, you were always given functions that were well-behaved in the sense that they were continuous over the interval you were integrating. Also, definite integrals were always between two finite values. Integration will work only under those conditions, i.e. you can't integrate over an interval that contains even one point where the function is discontinuous and, for definite integrals, the limits of integration must be finite. These are the requirements of the Fundamental Theorem of Calculus (FTC).

If you want a full lecture on this topic, we recommend this video.

Prof Leonard - Calculus 2 Lecture 7.6: Improper Integrals [2hr-48mins-30secs]

video by Prof Leonard

However, under certain conditions we can use limits to be able to integrate over discontinuous functions or where the limits of integration are not finite. These types of integrals are called improper integrals. The way to handle these kinds of integrals is to convert the integral to a finite integral on a continuous finite interval, integrate and then take the limit of the result. We can handle only one type of discontinuity at a time, so we may end up with multiple integrals but, as long as each of them are continuous and finite, that shouldn't be a problem.

Before we get into the details, let's watch a quick overview video that will help get us started.

Krista King Math - What makes an integral improper?! [4min-58secs]

video by Krista King Math

Before we get into the details of the techniques, here is a list of keys that you need to keep in mind as you learn these techniques. These keys will not make much sense at first but as you learn how to evaluate improper integrals, these keys will help keep you on track.

Improper Integral Keys

There are several keys to handling improper integrals.
1. We can handle only one 'problem' per integral.
2. The 'problem' must occur at one of the limits of integration or we need to break the integral to get it at one of the limits of integration.
3. Use a limit to set up the integral so that it is finite and continuous and then deal with the discontinuity after the integral is evaluated.

In summary, in all three cases, we use limits to rewrite the integral. We can do this because of the Limit Key. We can also have a combination of cases, in which case we rewrite the definite integral by breaking it into pieces so that we have only one case to handle for each integral.

The first thing we need to do is to determine the domain of the integrand and compare the domain with the limits of integration. Any points included in the limits of integration that are NOT in the domain are considered 'problem points'. Also, if one or either of the endpoints of is/are \(\pm\)infinity, they are also considered problems.

Let's start with the endpoints. The simplest case is where we have a continuous interval and one of the endpoints is infinite.

Case 1 - Infinite Interval

Case 1: Infinite Integral at One Endpoint - Problem and Example

\(\displaystyle{ \int_{a}^{\infty}{f(x) ~ dx} }\)

\(a\) is finite

\(f(x)\) is continuous on the interval \([a,\infty)\)

\(\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2} ~ dx} }\)

\(\displaystyle{ \int_{-\infty}^{b}{g(x) ~ dx} }\)

\(b\) is finite

\(g(x)\) is continuous on the interval \( (-\infty,b] \)

\(\displaystyle{ \int_{-\infty}^{-1}{\frac{1}{x^2} ~ dx} }\)

Special Improper Integral

\(\displaystyle{ \int_{1}^{\infty}{\frac{dx}{x^p}} = \left\{ \begin{array}{llr} \frac{1}{p-1} & & p \gt 1 \\ diverges & & p \leq 1 \end{array} \right. }\)

Notes
1. We chose the function \(1/x^2\) to demonstrate what we meant about being continuous on the interval of integration. Notice that as long as we stay away from \(x=0\), the function \(f(x)=1/x^2\) is continuous.
2. Also notice that we have a problem at only one endpoint in each integral. This is important, since we can handle only one problem at a time.
3. Using the techniques we know so far, we can't evaluate these integrals because it is required that the function be continuous and the definite integral can be evaluated only on a finite interval. So what do we do?

This is the case discussed in this video. He looks at the same function \(1/x^2\) and shows how to evaluate it with limits. He uses a graph to explain really well what is going on with this type of improper integral.

PatrickJMT - Improper Integral - Basic Idea and Example [6min-22secs]

video by PatrickJMT

So, to evaluate the integrals in the above table, we need these conditions.

Under These Conditions

\(a\), \(b\) and \(t\) are finite

\(f(x)\) is continuous on the interval \([a,\infty)\)

\(g(x)\) is continuous on the interval \( (-\infty,b] \)

We Can Write These Equations

\(\displaystyle{ \int_{a}^{\infty}{f(x) ~ dx} = }\) \(\displaystyle{\lim_{t \to \infty}{ \int_{a}^{t}{f(x) ~ dx} } }\)

\(\displaystyle{ \int_{-\infty}^{b}{g(x) ~ dx} = }\) \(\displaystyle{\lim_{t \to -\infty}{ \int_{t}^{b}{g(x) ~ dx} } }\)

Notice what we have done. We have replaced the infinity with a finite variable. This allows us to evaluate the integral. Once the integral is evaluated, we can then take the limit of the result. So, in essence, we have moved the 'problem' outside of the integral.

This is an important starting point since the upcoming cases rely on your understanding of this first case. Take a few minutes to work these practice problems before you go on.

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

Improper Integrals - Practice Problems Conversion

[1-1464] - [2-1465] - [3-994] - [4-995] - [5-1463] - [6-1466] - [7-1469] - [8-1471] - [9-1472]

[10-1475] - [11-1476] - [12-1820] - [13-993] - [14-1460] - [15-1473] - [16-1478] - [17-1470] - [18-1216]

[19-1477] - [20-1821] - [21-1461] - [22-1467] - [23-1468] - [24-1474] - [25-1479] - [26-1462] - [27-1480]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

Basic Problems

\(\displaystyle{ \int_{0}^{\infty}{ \frac{dx}{(x+1)^3} } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ \frac{dx}{(x+1)^3} } }\)

Final Answer

1/2

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ \frac{dx}{(x+1)^3} } }\)

Solution

Looking at the limits of integration, we have the upper limit at infinity. So we write the integral with a limit before integrating.

\(\displaystyle{ \lim_{b\to\infty}{ \int_{0}^{b}{ \frac{dx}{(x+1)^3} } } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \int_{0}^{b}{ (x+1)^{-3} ~dx } } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \left. \frac{(x+1)^{-2}}{-2} \right|_0^b } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{2(b+1)^2} - \frac{-1}{2(1)^2} \right] } = 0 + \frac{1}{2} = \frac{1}{2} }\)

Final Answer

1/2

close solution

\(\displaystyle{ \int_{1}^{\infty}{ 2^{-x} ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ 2^{-x} ~dx } }\)

Final Answer

\(\displaystyle{ \frac{1}{\ln 4} }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ 2^{-x} ~dx } }\)

Solution

First, let's write the integrand in terms of \(e\) so that integration is easier.

\(y=2^{-x}\)

\(\ln y = \ln 2^{-x}\)

\(\ln y = -x \ln 2\)

\( e^{\ln y} = e^{-x\ln 2} \)

\( y = e^{-x\ln 2} \)

So we will replace the integrand \(2^{-x}\) with \( e^{-x\ln 2} \).
Rewrite the integral with a limit and integrate.

\(\displaystyle{ \lim_{b\to\infty}{ \int_{1}^{b}{ e^{-x\ln 2} ~dx } } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \left. \frac{e^{-x\ln2}}{-\ln2} \right|_1^b } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{e^{b\ln2}\ln 2} - \frac{-1}{e^{\ln2}\ln 2} \right] } }\)

\(\displaystyle{ 0 + \frac{1}{2\ln 2} = \frac{1}{\ln 4} }\)

Final Answer

\(\displaystyle{ \frac{1}{\ln 4} }\)

close solution

\(\displaystyle{ \int_{1}^{\infty}{ x^2~dx } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ x^2~dx } }\)

Solution

1464 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{1}^{\infty}{ \frac{1}{x} dx } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ \frac{1}{x} dx } }\)

Solution

1465 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{2}^{\infty}{ \frac{dx}{x^5} dx } }\)

Problem Statement

\(\displaystyle{ \int_{2}^{\infty}{ \frac{dx}{x^5} dx } }\)

Solution

Make sure you understand that you do not plug in the infinite limit as if it was a number after evaluating the integral, like she does in this video. What is actually happening is that you are taking the limit. Before starting any work, the initial integral should be written
\(\displaystyle{ \int_{2}^{\infty}{ \frac{dx}{x^5} dx } = \lim_{b\to\infty}{\int_{2}^{b}{ \frac{dx}{x^5} dx } } }\)

994 solution video

video by Krista King Math

close solution

\(\displaystyle{ \int_{-\infty}^{0}{ e^{0.1x} dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{0}{ e^{0.1x} dx } }\)

Solution

Make sure you understand that you do not plug in the infinite limit as if it was number after evaluating the integral, like she does in this video. What is actually happening is that you are taking the limit. Before starting any work, the initial integral should be written
\(\displaystyle{ \int_{-\infty}^{0}{ e^{0.1x} dx } = \lim_{a\to-\infty}{\int_{a}^{0}{ e^{0.1x } dx } } }\)

995 solution video

video by Krista King Math

close solution

\(\displaystyle{ \int_{1}^{\infty}{ 2x^{-2} ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ 2x^{-2} ~dx } }\)

Solution

1463 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{\infty}{ 4e^{-2x} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ 4e^{-2x} dx } }\)

Solution

1466 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{\infty}{ 6e^{-3x} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ 6e^{-3x} dx } }\)

Solution

1469 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{\infty}{ \frac{1}{2x+5} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ \frac{1}{2x+5} dx } }\)

Solution

1471 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{-\infty}^{-3}{ \frac{2}{x^2} dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{-3}{ \frac{2}{x^2} dx } }\)

Solution

1472 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{\infty}{ 0.5e^xdx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ 0.5e^xdx } }\)

Solution

1475 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{1}^{\infty}{ \frac{8}{x^4} dx } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ \frac{8}{x^4} dx } }\)

Solution

1476 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{-\infty}^{0}{ \frac{dx}{3-4x} } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{0}{ \frac{dx}{3-4x} } }\)

Solution

1820 solution video

video by Krista King Math

close solution

Intermediate Problems

\(\displaystyle{ \int_{0}^{\infty}{ x e^{-3x} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ x e^{-3x} dx } }\)

Solution

This practice problem is solved in two consecutive videos.
Her notation here is not good and she makes a major jump (without explaining something) when evaluating one of the limits.
Her notation mistake is that she didn't write the integral as a limit. This may seem trivial but it isn't since notation will often be what will trip you up. So she should have started the problem by rewriting the integral as \(\displaystyle{\int_{0}^{\infty}{x e^{-3x} dx} = \lim_{b \to \infty}{ \int_{0}^{b}{x e^{-3x} dx}}}\) then worked the inside integral and evaluated the limit at the end.
Secondly, she evaluates the limit of one of the terms and said it was zero when actually it is indeterminate. The term was \(\displaystyle{\lim_{b \to \infty}{ \frac{-be^{-3b}}{3}}}\).
Since she didn't write out the limit, this may be hard to see but it is the first term when she plugged in \( \infty \). The term she got was \(\displaystyle{\frac{-\infty e^{-3\infty}}{3}}\) which can be rewritten as \(\displaystyle{\frac{-\infty }{3e^{3\infty}} = \frac{-\infty}{\infty}}\) and is clearly indeterminate. To evaluate the limit, we need to use L'Hopitals rule as follows
\(\displaystyle{ \lim_{b \to \infty}{ \frac{-be^{-3b}}{3} } = }\) \(\displaystyle{ \lim_{b \to \infty}{ \frac{-b}{3e^{3b}} } = }\) \(\displaystyle{ \lim_{b \to \infty}{ \frac{-1}{9e^{3b}} } = \frac{-1}{\infty} = 0 }\)

So, that is where the zero comes from in that term. She either got lucky or she just didn't show the steps, neither of which should be taken for granted. Other than these mistakes, this is a good example.

993 solution video

video by Krista King Math

993 solution video

video by Krista King Math

close solution

\(\displaystyle{ \int_{0}^{\infty}{ se^{-5s} ~ds } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ se^{-5s} ~ds } }\)

Solution

1460 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int_{-\infty}^{2}{ \frac{3}{x^2+1} dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{2}{ \frac{3}{x^2+1} dx } }\)

Solution

1473 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{\infty}{ \cos(x) ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\infty}{ \cos(x) ~dx } }\)

Solution

1478 solution video

video by MIT OCW

close solution

\(\displaystyle{ \int_{-\infty}^0{ x e^x ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^0{ x e^x ~dx } }\)

Final Answer

\(-1\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^0{ x e^x ~dx } }\)

Solution

2223 solution video

video by Michel vanBiezen

Final Answer

\(-1\)

close solution

So, you should now be comfortable evaluating integrals where the integrand is continuous on the entire interval of integration and with one of the endpoints going to infinity or negative infinity. But what do you do if both limits of integration contain infinity, i.e. the lower limit is negative infinity AND the upper limit is infinity. The trick is to break the integral at a convenient point and then you have two integrals. For example, if we have \(\displaystyle{ \int_{-\infty}^{\infty}{ x ~ dx} }\), we can break the integral at any point \(a\) to get integrals.
\(\displaystyle{ \int_{-\infty}^{\infty}{ x ~ dx} = }\) \(\displaystyle{ \int_{-\infty}^{0}{ x } + }\) \(\displaystyle{ \int_{0}^{\infty}{ x ~ dx} }\)
We chose zero in the above example but you can choose any point or even use \(a\). Then we use the technique we learned above to evaluate each integral, adding the results together to get our final answer.

Here are some practice problems to try.

\(\displaystyle{ \int_{-\infty}^{\infty}{ -3x^4 ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{\infty}{ -3x^4 ~dx } }\)

Solution

1470 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{-\infty}^{\infty}{ x^2 e^{-x^3} ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{\infty}{ x^2 e^{-x^3} ~dx } }\)

Solution

1216 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{1}{1+x^2} dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{1}{1+x^2} dx } }\)

Solution

1477 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{x^2}{9+x^6} dx } }\)

Problem Statement

\(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{x^2}{9+x^6} dx } }\)

Solution

1821 solution video

video by Krista King Math

close solution

Case 2 - Discontinuity at an Endpoint

This case occurs when we have a continuous function within a finite interval but the function is discontinuous at one of the endpoints. For example, we might have an integral like \(\displaystyle{ \int_{0}^{1}{ \frac{1}{x} ~ dx} }\). In this case, \(f(x)=1/x\) is discontinuous at one of the endpoints, \(x=0\). So we can't integrate. However, we can use the same technique above that we used for an infinite interval and use limits. It looks something like this.

Case 2: Finite Integral with a Discontinuity at One Endpoint

\(\displaystyle{ \int_{a}^{b}{f(x) ~ dx} }\)

\(a\) and \(b\) are finite

\(f(x)\) is continuous on the interval \([a,b)\) but discontinuous at \(x=b\)

\(\displaystyle{\lim_{t \to b}{ \int_{a}^{t}{f(x) ~ dx} } }\)

\(\displaystyle{ \int_{a}^{b}{g(x) ~ dx} }\)

\(a\) and \(b\) are finite

\(g(x)\) is continuous on the interval \( (a,b] \) but discontinuous at \(x=a\)

\(\displaystyle{\lim_{t \to a}{ \int_{t}^{b}{g(x) ~ dx} } }\)

Notice that we do the same thing as in Case 1, i.e. we replace the 'problem' point with a limit. Okay, try these practice problems before going on.

\(\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } }\)

Problem Statement

\(\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } }\)

Final Answer

4

Problem Statement

\(\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } }\)

Solution

\(\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } = \lim_{a\to 0^+}{ \int_a^8{ \frac{dx}{\sqrt{2x}} } } }\)

\(\displaystyle{ \int{\frac{dx}{\sqrt{2x}}} }\)

\(\displaystyle{ \frac{1}{\sqrt{2}} \int{ x^{-1/2}~dx } }\)

\(\displaystyle{ \frac{1}{\sqrt{2}} \frac{x^{1/2}}{1/2} }\)

\(\displaystyle{ \sqrt{2x} }\)

\(\displaystyle{ \lim_{a\to 0^+}{ \left[ \sqrt{2x} \right]_a^8 } }\)

\(\displaystyle{ \lim_{a\to 0^+}{ [ \sqrt{16} - \sqrt{2a} ] } = 4 }\)

Final Answer

4

close solution

\(\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }\)

Problem Statement

\(\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }\)

Hint

Use trig substitution.

Problem Statement

\(\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }\)

Final Answer

\(\pi/2\)

Problem Statement

\(\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }\)

Hint

Use trig substitution.

Solution

We start by rewriting the integral using a limit.
\(\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } = \lim_{b\to3^-}{ \int_0^b{ \frac{dx}{\sqrt{9-x^2}} } } }\)
Now we will evaluate the indefinite integral
\(\displaystyle{ \int{ \frac{dx}{\sqrt{9-x^2}} } }\)
Use trig substitution with \(x=3\sin\theta \to dx = 3\cos\theta~d\theta\).
\(\displaystyle{ \int{ \frac{3\cos\theta~d\theta}{\sqrt{9-9\sin^2\theta}} } = \int{ \frac{\cos\theta}{\cos\theta}~d\theta } = \int{1~d\theta} = \theta = \arcsin(x/3) }\)
\(\displaystyle{ \lim_{b\to3^-}{ [ \arcsin(x/3)|_0^b ] } = \lim_{b\to3^-}{ [ \arcsin(b/3) - \arcsin(0/3) ] } = \arcsin 1 = \pi/2 }\)

Final Answer

\(\pi/2\)

close solution

\(\displaystyle{ \int_{3}^{4}{ \frac{dx}{(x-3)^{3/2}} } }\)

Problem Statement

\(\displaystyle{ \int_{3}^{4}{ \frac{dx}{(x-3)^{3/2}} } }\)

Final Answer

The integral diverges.

Problem Statement

\(\displaystyle{ \int_{3}^{4}{ \frac{dx}{(x-3)^{3/2}} } }\)

Solution

Comparing the limits of integration to the integrand, we can see that there is a discontinuity at the lower limit. This is the only problem point. So we rewrite the integral as
\(\displaystyle{ \lim_{a \to 3^+}{ \int_{a}^{4}{ \frac{dx}{(x-3)^{3/2}} } } }\)
We now evaluate the indefinite integral.

\(\displaystyle{ \int{ \frac{dx}{(x-3)^{3/2}} } }\)

\( \int{ (x-3)^{-3/2} ~dx } \)

\( (x-3)^{-1/2} / (-1/2) \)

\( -2(x-3)^{-1/2} \)

Now plug this result back into the limit and evaluate.

\(\displaystyle{ \lim_{a \to 3^+}{ [-2(x-3)^{-1/2}]_a^4 } }\)

\(\displaystyle{ \lim_{a \to 3^+}{ [-2(4-3)^{-1/2} + 2(a-3)^{-1/2}] } }\)

\( -2 + \infty = \infty \)

Final Answer

The integral diverges.

close solution

\(\displaystyle{ \int_{-3}^{1}{ \frac{dx}{(2x+6)^{2/3}} } }\)

Problem Statement

\(\displaystyle{ \int_{-3}^{1}{ \frac{dx}{(2x+6)^{2/3}} } }\)

Final Answer

3

Problem Statement

\(\displaystyle{ \int_{-3}^{1}{ \frac{dx}{(2x+6)^{2/3}} } }\)

Solution

At \(x=-3\), the denominator of the integrand is zero. This is the only problem point and it occurs at the lower limit of integration. So we rewrite the integral as \(\displaystyle{ \lim_{a \to -3^+}{ \int_{a}^{1}{ \frac{dx}{(2x+6)^{2/3}} } } }\)
Now we evaluate the indefinite integral.

\(\displaystyle{ \int{ \frac{dx}{(2x+6)^{2/3}} } }\)

Use integration by substitution.

\(u=2x+6 \to du=2dx\)

\(\displaystyle{ \int{ \frac{du}{2 u^{2/3}} } }\)

\(\displaystyle{ \frac{1}{2} \int{ u^{-2/3} ~du } }\)

\(\displaystyle{ \frac{1}{2} \frac{u^{1/3}}{1/3} }\)

\(\displaystyle{ \frac{3}{2} u^{1/3} }\)

\(\displaystyle{ \frac{3}{2} (2x+6)^{1/3} }\)

Now plug this result back into the limit and evaluate.

\(\displaystyle{ \lim_{a \to -3^+}{ \frac{3}{2} [(2x+6)^{1/3}]_a^1 } }\)

\(\displaystyle{ \frac{3}{2} \lim_{a \to -3^+}{ [(2+6)^{1/3} - (2a+6)^{1/3}] } }\)

\(\displaystyle{ \frac{3}{2} [(8)^{1/3} - (2(-3)+6)^{1/3}] = 3 }\)

Final Answer

3

close solution

\(\displaystyle{ \int_{0}^{\pi/2}{ \tan\theta ~ d\theta } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{\pi/2}{ \tan\theta ~ d\theta } }\)

Final Answer

The integral diverges.

Problem Statement

\(\displaystyle{ \int_{0}^{\pi/2}{ \tan\theta ~ d\theta } }\)

Solution

Looking at the integrand, we know that tangent is not defined at \(\pi/2\). So that is the only problem point. We rewrite the integral with the limit and solve.

\(\displaystyle{ \lim_{b\to\pi/2^-}{ \int_{0}^{b}{ \tan\theta ~ d\theta } } }\)

\(\displaystyle{ \lim_{b\to\pi/2^-}{ \left[ -\ln|\cos\theta| \right]_0^b } }\)

\(\displaystyle{ \lim_{b\to\pi/2^-}{ \left[ -\ln|\cos b| + \ln|\cos 0| \right] } }\)

\( -\ln|\cos \pi/2| + \ln|1| \)

\( -\ln|0| + 0 = +\infty \)

Final Answer

The integral diverges.

close solution

\(\displaystyle{ \int_{-1}^{0}{ \frac{1}{x^2} dx } }\)

Problem Statement

\(\displaystyle{ \int_{-1}^{0}{ \frac{1}{x^2} dx } }\)

Solution

1461 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int_{0}^{6}{\frac{1}{ x^{0.8}} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{6}{\frac{1}{ x^{0.8}} dx } }\)

Solution

1467 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{3}{ \frac{1}{x^{1.2}} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{3}{ \frac{1}{x^{1.2}} dx } }\)

Solution

1468 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{5}^{8}{ \frac{2}{\sqrt[3]{x-5}} dx } }\)

Problem Statement

\(\displaystyle{ \int_{5}^{8}{ \frac{2}{\sqrt[3]{x-5}} dx } }\)

Solution

1474 solution video

video by MIP4U

close solution

\(\displaystyle{ \int_{0}^{1}{ \frac{\ln(x)}{x^{1/2}} dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{1}{ \frac{\ln(x)}{x^{1/2}} dx } }\)

Solution

1479 solution video

video by MIT OCW

close solution

\(\displaystyle{ \int_{-1}^1{ \frac{x}{x^2+2x+1} ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{-1}^1{ \frac{x}{x^2+2x+1} ~dx } }\)

Final Answer

The integral diverges.

Problem Statement

\(\displaystyle{ \int_{-1}^1{ \frac{x}{x^2+2x+1} ~dx } }\)

Solution

\(\displaystyle{ \frac{x}{x^2+2x+1} = \frac{x}{(x+1)^2} }\)
The domain of the integrand is all real numbers except for \(x=-1\), which is the lower limit of integration. This is the only problem point. So we rewrite the integral as
\(\displaystyle{ \lim_{a\to-1^+}{ \int_{a}^1{ \frac{x}{x^2+2x+1} ~dx } } }\)
Now we integrate the indefinite integral.

\(\displaystyle{ \int{ \frac{x}{x^2+2x+1} ~dx } }\)

\( u=x+1 \to du=dx \) and \(x=u-1\)

\(\displaystyle{ \int{ \frac{u-1}{u^2} ~du } }\)

\(\displaystyle{ \int{ \frac{1}{u} - \frac{1}{u^2} ~du } }\)

\(\displaystyle{ \ln|u| - \frac{u^{-1}}{-1} }\)

\(\displaystyle{ \ln|u| + \frac{1}{u} }\)

\(\displaystyle{ \ln|x+1| + \frac{1}{x+1} }\)

Now plug this result back into the limit.

\(\displaystyle{ \lim_{a\to-1^+}{ \left[ \ln|x+1| + \frac{1}{x+1} \right]_a^1 } }\)

\(\displaystyle{ \lim_{a\to-1^+}{ \left[ \ln 2 + 1/2 - \ln|a+1| - 1/(a+1) \right] } }\)

\( \ln 2 + 1/2 - (-\infty) - \infty = \infty - \infty \)

Since \( \infty - \infty \) is indeterminate, we need to drop back and see where those terms came from and evaluate them differently. The second to the last step shows us that they come from the last two terms. So we will evaluate them separately using L'Hopitals Rule.
Note: Since \(a \gt -1\), we can write \(|a+1|\) without the absolute values signs, i.e. \(|a+1| = (a+1)\).

\(\displaystyle{ \lim_{a\to-1^+}{ \left[ -\ln(a+1) - 1/(a+1) \right] } }\)

\(\displaystyle{ -\lim_{a\to-1^+}{ \left[ \ln(a+1) + 1/(a+1) \right] } }\)

\(\displaystyle{ -\lim_{a\to-1^+}{ \left[ \frac{(a+1)\ln(a+1)+1}{a+1} \right] } }\)

Now apply L'Hopitals Rule

\(\displaystyle{ -\lim_{a\to-1^+}{ \left[ \frac{(a+1)[1/(a+1)] + \ln(a+1)(1) }{1} \right] } }\)

\(\displaystyle{ -\lim_{a\to-1^+}{ \left[ 1+\ln(a+1) \right] } }\)

\( -[ 1 + (-\infty) ] = \infty \)

Combining results, we get \(\ln 2 + 1/2 + \infty = \infty \). So the integral diverges.

Final Answer

The integral diverges.

close solution

\(\displaystyle{ \int_{0}^{1}{ x^2\ln(1/x) ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{1}{ x^2\ln(1/x) ~dx } }\)

Final Answer

1/9

Problem Statement

\(\displaystyle{ \int_{0}^{1}{ x^2\ln(1/x) ~dx } }\)

Solution

Comparing the limits of integration to the integrand, we can see that the only problem point is the lower limit, \(x=0\). So we rewrite the limit as follows.
\(\displaystyle{ \lim_{a\to0^+}{ \int_{a}^{1}{ x^2\ln(1/x) ~dx } } }\)
We will now evaluate the indefinite integral.

\( \int{ x^2\ln(1/x) ~dx } \)

\( \int{ -x^2\ln(x) ~dx } \)

Use integration by parts.

\(u=\ln x \to du=(1/x)dx\)

\(dv = x^2~dx \to v=x^3/3\)

\(\displaystyle{ - \left[ \frac{x^3}{3} \ln x - \int{ \frac{x^3}{3} \frac{1}{x} ~dx } \right] }\)

\(\displaystyle{ - \frac{x^3}{3} \ln x + \frac{1}{3} \int{ x^2 ~dx } }\)

\(\displaystyle{ - \frac{x^3}{3} \ln x + \frac{1}{3} \frac{x^3}{3} }\)

\(\displaystyle{ \frac{x^3}{9} [ 1 - 3\ln x ] }\)

Now we take this result and plug it back into the limit.

\(\displaystyle{ \lim_{a\to0^+}{ \left. \frac{x^3}{9} ( 1 - 3\ln x ) \right|_a^1 } }\)

\(\displaystyle{ \lim_{a\to0^+}{ \left[ \frac{1}{9} ( 1 - 3\ln 1 ) - \frac{a^3}{9} ( 1 - 3\ln a ) \right] } }\)

\(\displaystyle{ \left[ \frac{1}{9} - \frac{0^3}{9} ( 1 - 3\ln 0 ) \right] }\)

\(\displaystyle{ \left[ \frac{1}{9} - 0 \cdot \infty \right] }\)

In the last line above, \(0 \cdot \infty\) is indeterminate. So we have to drop back and see where they came from so that we can evaluate the limit differently. We need to look at this limit.

\(\displaystyle{ \lim_{a\to0^+}{ a^3\ln a } }\)

To use L'Hopitals Rule, we need a fraction.

\(\displaystyle{ \lim_{a\to0^+}{ \frac{\ln a}{a^{-3}} } }\)

Apply L'Hopitals Rule.

\(\displaystyle{ \lim_{a\to0^+}{ \frac{1/a}{-3a^{-4}} } }\)

\(\displaystyle{ \lim_{a\to0^+}{ \frac{-1}{3} a^3 } = 0 }\)

Combining our results, we have \(1/9-0 = 1/9\)

Final Answer

1/9

close solution

So far, so good. Now, there is only one minor twist to these two cases that is easily handled. If you have a 'problem' (a discontinuity or infinity) at both ends, you just break the integral into two integrals at a convenient point. We know from basic integration that \(\displaystyle{ \int_{a}^{b}{f(x)~dx} = }\) \(\displaystyle{ \int_{a}^{k}{f(x)~dx} + \int_{k}^{b}{f(x)~dx} }\), right? So we just do the same thing with the improper integral and handle each case and limit separately. For example, we might want to evaluate \(\displaystyle{ \int_{0}^{\infty}{\frac{1}{x}~dx} }\), so we just split the integral into two and handle each end separately. We can break it anywhere in the interval, I will choose \(2\). We end up with
\(\displaystyle{ \int_{0}^{\infty}{\frac{1}{x}~dx} = }\) \(\displaystyle{ \int_{0}^{2}{\frac{1}{x}~dx} + \int_{2}^{\infty}{\frac{1}{x}~dx} = }\) \(\displaystyle{ \lim_{a \to 0}{\int_{a}^{2}{\frac{1}{x}~dx} } + \lim_{b \to \infty}{ \int_{2}^{b}{\frac{1}{x}~dx} } }\)

Case 3 - Discontinuity Inside the Interval

Cases 1 and 2 above were very similar and dealing with them used the similar technique of taking the limit of one of ends of the interval. Now, Case 3 may seem very different but handling it is very straightforward and, if you understand Cases 1 and 2, you will be able to deal with Case 3 easily. Case 3 involves discontinuities INSIDE the interval, i.e. not at one of the endpoints or limits of integration. To handle this case, you just break the integral again (like you did in Case 2) but this time you break it AT THE POINT OF DISCONTINUITY (not at just any point). When you do, you end up with something like this.

Under These Conditions

\(a\) and \(b\) are finite

\(f(x)\) is continuous on the interval \([a,b]\) but discontinuous at \(x=c\), where \(a < c < b\)

This Equation Holds

\(\displaystyle{ \int_{a}^{b}{f(x) ~ dx} = }\) \(\displaystyle{ \int_{a}^{c}{f(x) ~ dx} + \int_{c}^{b}{f(x) ~ dx} }\)

Since \(f(x)\) is discontinuous at \(x=c\), you now have two integrals that fit Case 2. So you can use those techniques to evaluate the integrals. Pretty cool, eh? An example might be \(\displaystyle{ \int_{-1}^{1}{1/x~dx} = \int_{-1}^{0}{1/x~dx} + \int_{0}^{1}{1/x~dx} }\).

Additional Situations You Will See
1. Another minor twist might be that you have an infinite limit at one or both ends AND a discontinuity in the interval. In this case, break the integral at the discontinuity and drop back to Cases 1 and 2 until you have only one 'problem' in each integral. You probably also know what to do with multiple discontinuities, right? Just break the integral at each discontinuity.
2. You may also have more than one discontinuity in the interval. As long as you have a finite number of discontinuties, just break the integral into more than two integrals at each discontinuity. Make sure that you end up with only one 'problem' point for each integral.

Okay, try these practice problems before going on.

\(\displaystyle{ \int_{-1}^{1}{ x^{-2/3} ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{-1}^{1}{ x^{-2/3} ~dx } }\)

Solution

1480 solution video

video by MIT OCW

close solution

\(\displaystyle{ \int_{-1}^{1}{ \frac{e^x}{e^x-1} ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{-1}^{1}{ \frac{e^x}{e^x-1} ~dx } }\)

Solution

In the video, he incorrectly evaluates \(\ln|e^x-1|\) at \(x=-1\) as \(\ln(2)\). It should be \(\ln|e^{-1}-1| = \ln|1/e-1|\). This could be simplified to \(\ln|1-e|-1\). However, this mistake does not change his final answer, which is correct, that the integral evaluates to \(-\infty\) and therefore the integral diverges.

1462 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } }\)

Problem Statement

\(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } }\)

Final Answer

\(3(2^{1/3})+6\)

Problem Statement

\(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } }\)

Solution

The domain of the integrand is all reals except for \(x=3\). The discontinuity is within the interval of integration, so we need to break the integral into two integrals at that point.
\(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } = \int_1^{3}{ \frac{dx}{(x-3)^{2/3}} } + \int_3^{11}{ \frac{dx}{(x-3)^{2/3}} } }\)
Now we rewrite the integrals with limits, so that we can integrate using the FTC.
\(\displaystyle{ \lim_{b\to3^-}{ \left[ \int_1^{b}{ \frac{dx}{(x-3)^{2/3}} } \right] } + \lim_{a\to3^+}{ \left[ \int_a^{11}{ \frac{dx}{(x-3)^{2/3}} } \right] } }\)
To save space, we will integrate the indefinite integral.

\(\displaystyle{ \int{ \frac{dx}{(x-3)^{2/3}} } }\)

\( u=x-3 \to du=dx \)

\( \int{ (u)^{-2/3} ~du } \)

\(\displaystyle{ \frac{u^{1/3}}{1/3} }\)

\( 3(x-3)^{1/3} \)

Now, plug this result back into the limit equations.

\(\displaystyle{ 3\lim_{b\to3^-}{ (x-3)^{1/3}|_1^b } + 3\lim_{a\to3^+}{ (x-3)^{1/3}|_a^{11} } }\)

\(\displaystyle{ 3\lim_{b\to3^-}{ [ (b-3)^{1/3} - (1-3)^{1/3} ] } + 3\lim_{a\to3^+}{ [ (11-3)^{1/3} - (a-3)^{1/3} ] } }\)

\(\displaystyle{ -3(-2)^{1/3}+3(8)^{1/3} = 3(2^{1/3})+6 }\)

Final Answer

\(3(2^{1/3})+6\)

close solution

How To Apply These Techniques to Multiple Discontinuities

So, those are the only discontinuities that you had to deal with. However, it helps to have a set of logical steps to follow to make sure you have all possible cases covered. Here are our suggestions.
This may seem like a lot of steps but some may not be required depending on your integral. These steps cover the most involved integration possible.

1. Make a list of all the 'problem' points, which include infinite limits of integration and discontinuities, points that are not in the domain of the integrand.

2. Discard any discontinuities that are outside the limits of integration.

3. List the 'problem' points left to right (or top to bottom) in increasing order with the endpoints of the integration at each end (the endpoints may or may not be 'problems'; so keep track of that information).

4. Break the integral at each discontinuity.

5. Break the integral into pieces between discontinuities so that each integral has only one 'problem' point at one of the ends of the interval.

6. Set up each integral individually using limits. (It is best to use different limit variables for each integral but many instructors do not require it.)

7. Evaluate the integral once in a separate area on your paper.

8. Use your result from step 7 in each limit and evaluate them separately.

9. Add your answers together. In this step, you may get one or several indeterminate forms. If you do, drop back to step 8 and determine where they come from. Combine those results before taking the limit. This is something you will get a feel for as you get experience.
Note - - In this last step, you may get an indeterminate form \(\infty - \infty\). This requires special handling discussed in the indeterminate differences section on the indeterminate forms page.

Okay, let's work some practice problems that may require multiple techniques. These are the types of problems that you will most likely see on your exam.

Evaluate \(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } }\)

Final Answer

\(\displaystyle{ \frac{1}{2048} }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } }\)

Solution

For this integral, there is only one problem point, the upper limit of integration. The discontinuity at \(x=-1\) can be ignored since it is not within the limits of integration. So we have \(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } = \lim_{b\to\infty}{ \int_{1}^{b}{ \frac{dx}{(x+1)^9} } } }\).
We will now integrate the indefinite integral \(\displaystyle{ \int{ \frac{dx}{(x+1)^9} } }\) and plug the result back into the definite integral and limit.

\(\displaystyle{ \int{ \frac{dx}{(x+1)^9} } }\)

Use integration by substitution.

\( u = x+1 \to du = dx \)

\(\displaystyle{ \int{ \frac{du}{u^9} } }\)

\(\displaystyle{ \int{ u^{-9} ~du } }\)

\(\displaystyle{ \frac{u^{-8}}{-8} }\)

\(\displaystyle{ \frac{-1}{8(x+1)^8} }\)

Plugging this result back into the limit, we have

\(\displaystyle{ \lim_{b\to\infty}{ \left[ \left. \frac{-1}{8(x+1)^8} \right|_1^b \right] } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{8(b+1)^8} + \frac{1}{8(2)^8} \right] } }\)

\(\displaystyle{ 0 + \frac{1}{2^{11}} = \frac{1}{2048} }\)

Final Answer

\(\displaystyle{ \frac{1}{2048} }\)

close solution

Set up integrals and limits equivalent to \(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{dx}{x(x+1)^9} } }\) so that we can use the FTC but do not evaluate the integrals.

Problem Statement

Set up integrals and limits equivalent to \(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{dx}{x(x+1)^9} } }\) so that we can use the FTC but do not evaluate the integrals.

Hint

FTC = Fundamental Theorem of Calculus

Problem Statement

Set up integrals and limits equivalent to \(\displaystyle{ \int_{-\infty}^{\infty}{ \frac{dx}{x(x+1)^9} } }\) so that we can use the FTC but do not evaluate the integrals.

Hint

FTC = Fundamental Theorem of Calculus

Solution

First we need to find the domain of the integrand. The problem points occur where the denominator is zero, i.e. at \(x=0\) and \(x=-1\). So the domain is all reals except for these two points.
The upper and lower limits of integration are infinite, so those are problems too. Here is a list of problem points, in numerical order.

\(-\infty\)

\(x=-1\)

\(x=0\)

\(+\infty\)

Now we compare the discontinuities against the interval of integration, \((-\infty,+\infty)\). Both \(x=-1\) and \(x=0\) are inside this interval, so we need to keep both of them. Now we can start setting up the integrals.

First we handle case 3, discontinuities inside the interval. We break the integral at each discontinuity. To make these easier to write, we will write the integrand as \(f(x)\), i.e. \(\displaystyle{ f(x)=\frac{1}{x(x+1)^9} }\)

\(\displaystyle{ \int_{-\infty}^{-1}{ f(x)~dx } + \int_{-1}^{0}{ f(x)~dx } + \int_{0}^{\infty}{ f(x)~dx } }\)

Okay, so at this point all the problem points are at the ends of integration limits and, within the limits of integration for each integral, the function is continuous. Now we will handle the end points of each integral separately.

\(\displaystyle{ \int_{-\infty}^{-1}{ f(x)~dx } }\)

Since both endpoints are problem points, we need to break the integral into two integrals at any point between \(-\infty\) and \(-1\). We choose \(-5\).

\(\displaystyle{ \int_{-\infty}^{-5}{ f(x)~dx } + \int_{-5}^{-1}{ f(x)~dx } }\)

Now each integral has only one problem point.

Repeat the same procedure on the other two integrals.

\(\displaystyle{ \int_{-1}^{0}{ f(x)~dx } = \int_{-1}^{-1/2}{ f(x)~dx } + \int_{-1/2}^{0}{ f(x)~dx } }\)

\(\displaystyle{ \int_{0}^{\infty}{ f(x)~dx } = \int_{0}^{17}{ f(x)~dx } + \int_{17}^{\infty}{ f(x)~dx } }\)

Let's put our results together to see what we have at this point. \(\displaystyle{ \int_{-\infty}^{\infty}{ f(x)~dx } = \int_{-\infty}^{-5}{ f(x)~dx } + \int_{-5}^{-1}{ f(x)~dx } + \int_{-1}^{-1/2}{ f(x)~dx } + \int_{-1/2}^{0}{ f(x)~dx } + \int_{0}^{17}{ f(x)~dx } + \int_{17}^{\infty}{ f(x)~dx } }\)
We have six integrals with each integral having only one problem point and each problem point occurs at one of the limits of integration. We are now ready to set up limits as follows.

\(\displaystyle{ \int_{-\infty}^{\infty}{ f(x)~dx } = \lim_{a\to-\infty}{ \int_{a}^{-5}{ f(x)~dx } } + \lim_{b\to-1^-}{ \int_{-5}^{b}{ f(x)~dx } } + \lim_{c\to-1^+}{ \int_{c}^{-1/2}{ f(x)~dx } } + \lim_{d\to0^-}{ \int_{-1/2}^{d}{ f(x)~dx } } + \lim_{y\to0^+}{ \int_{y}^{17}{ f(x)~dx } } + \lim_{z\to\infty}{ \int_{17}^{z}{ f(x)~dx } } }\)
A few comments are in order.
1. Notice that we used different variables in each limit. This may or may not be required by your instructor but it is highly recommended by us since it will keep you from getting confused when evaluating the limits.
2. We have left \(f(x)\) in the answer, but your instructor may require you to write out the integrand within each limit. However, as an instructor, I would be okay with this answer as long as you also write what \(f(x)\) was equal to along with your answer as shown below.
3. As we mentioned in the solution, you may choose values other than \(-5\), \(-1/2\) and \(17\). As long as you meet the requirement that the point is within each limit of integration, your answer is correct.
4. As always, check with your instructor to see what they require.

Final Answer

\(\displaystyle{ \int_{-\infty}^{\infty}{ f(x)~dx } = \lim_{a\to-\infty}{ \int_{a}^{-5}{ f(x)~dx } } + \lim_{b\to-1^-}{ \int_{-5}^{b}{ f(x)~dx } } + \lim_{c\to-1^+}{ \int_{c}^{-1/2}{ f(x)~dx } } + \lim_{d\to0^-}{ \int_{-1/2}^{d}{ f(x)~dx } } + \lim_{y\to0^+}{ \int_{y}^{17}{ f(x)~dx } } + \lim_{z\to\infty}{ \int_{17}^{z}{ f(x)~dx } } }\)

where \(\displaystyle{ f(x)=\frac{1}{x(x+1)^9} }\)

close solution

\(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt[3]{x-1}} } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt[3]{x-1}} } }\)

Final Answer

The integral diverges.

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt[3]{x-1}} } }\)

Solution

Looking at the limits of integration and comparing them with the integrand, it should be obvious that we have a problem at both ends. We have infinity at the upper limit and a discontinuity at the lower limit. However, the integrand is continuous in the interval \((1,\infty)\). So the first step is to break the integral into two integrals at any point inside the interval. This will give us two integrals with only one problem point each, allowing us to write limits for both. We choose to break the interval at 17.

\(\displaystyle{ \int_{1}^{17}{ \frac{dx}{\sqrt[3]{x-1}} } + \int_{17}^{\infty}{ \frac{dx}{\sqrt[3]{x-1}} } }\)

\(\displaystyle{ \lim_{a\to1^+}{ \int_{a}^{17}{ \frac{dx}{\sqrt[3]{x-1}} } } + \lim_{b\to\infty}{ \int_{17}^{b}{ \frac{dx}{\sqrt[3]{x-1}} } } }\)

Now we will evaluate the indefinite integral.

\(\displaystyle{ \int{ \frac{dx}{\sqrt[3]{x-1}} } }\)

\( \int{ (x-1)^{-1/3}~dx } \)

\(\displaystyle{ \frac{(x-1)^{2/3}}{2/3} }\)

\(\displaystyle{ \frac{3}{2}(x-1)^{2/3} }\)

Next, we plug this result back into the both limits and evaluate.

\(\displaystyle{ \lim_{a\to1^+}{ \left[ \frac{3}{2}(x-1)^{2/3} \right]_a^{17} } + \lim_{b\to\infty}{ \left[ \frac{3}{2}(x-1)^{2/3} \right]_{17}^{b} } }\)

\(\displaystyle{ \frac{3}{2} \lim_{a\to1^+}{ \left[ (17-1)^{2/3} - (a-1)^{2/3} \right] } + \frac{3}{2} \lim_{b\to\infty}{ \left[ (b-1)^{2/3} - (17-1)^{2/3} \right] } }\)

\(\displaystyle{ \frac{3}{2} \left[ (16)^{2/3} - 0 \right] + \frac{3}{2} \left[ \infty - (16)^{2/3} \right] = \infty }\)

Final Answer

The integral diverges.

close solution

\(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{x^2(x+1)} } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{x^2(x+1)} } }\)

Final Answer

\( 1 - \ln 2 \)

Problem Statement

\(\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{x^2(x+1)} } }\)

Solution

Looking at the limits of integration, we can see that the upper limit is a problem since it goes to infinity. Before we jump in, let's look at the domain of the integrand. The domain is all real number except for the values that make the denominator zero, in this case, \(x=0\) and \(x=-1\). So those points are potential problems as well. However, comparing these points to the limits of integration shows us that these points are outside the integration interval and, therefore, do not impact our problem. So even though it initially looked like there were several problem points, it turns out there is only one, the upper limit.
So now we will rewrite the integral with a limit.
\(\displaystyle{ \lim_{b\to\infty}{ \int_{1}^{b}{ \frac{dx}{x^2(x+1)} } } }\)
Now we will evaluate the indefinite integral.

\(\displaystyle{ \int{ \frac{dx}{x^2(x+1)} } }\)

Use partial fractions to expand the integrand.

\(\displaystyle{ \int{ \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{x+1} ~ dx } }\)

\(\displaystyle{ -\ln|x| + \frac{x^{-1}}{-1} + \ln|x+1| }\)

Plug this back into the limit.

\(\displaystyle{ \lim_{b\to\infty}{ \left[ -\ln|x| - 1/x + \ln|x+1| \right]_1^b } }\)

\(\displaystyle{ \lim_{b\to\infty}{ \left[ -\ln|b| - 1/b + \ln|b+1| \right] - \left[ -\ln|1| - 1/1 + \ln|1+1| \right] } }\)

\(\displaystyle{ \left[ -\infty - 0 + \infty \right] - \left[ -0 - 1 + \ln 2 \right] }\)

In the first set of brackets we have \(\infty-\infty\), which is indeterminate. So we need to determine where these came from and evaluate the limit differently. We have
\(\displaystyle{ \lim_{b\to\infty}{ [ -\ln|b| + \ln|b+1| ] } }\)
Combining the natural log terms, we have
\(\displaystyle{ \lim_{b\to\infty}{ \ln\left| \frac{b+1}{b} \right| } }\)
\(\displaystyle{ \lim_{b\to\infty}{ \ln\left| 1 + \frac{1}{b} \right| } = 0 }\)
So that leaves us with \(-[-1+\ln 2] = 1 - \ln 2 \)

Final Answer

\( 1 - \ln 2 \)

close solution
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