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The idea of using partial fractions to integrate is to convert the integrand into a form that we can integrate using techniques we've learned so far. As you know by now, integration is much more difficult that differentiation. When we have a fraction with polynomials in the denominator that can be factored, we can sometimes separate the fraction into two or more fractions that we use substitution or another known technique to integrate.

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

Prof Leonard - Calculus 2 Lecture 7.4: Integration By Partial Fractions [2hrs-55mins-42secs]

video by Prof Leonard

Partial Fraction Expansion Precalculus Review

The problem most students have when using this technique is recalling how to do partial fraction expansion that they learned in precalculus. So the first section on this page contains a lot of information to refresh your skills on this technique.

Here are some panels containing a review of partial fraction expansion (decomposition) from precalculus. For practice problems and more discussion, see the precalculus partial fraction expansion page.

Steps

As we present each of these steps, we will work through an example to show you how they work.
Example To Show Steps: Expand the fraction \(\displaystyle{ \frac{1}{x^2+3x+2} }\) using the technique of partial fraction expansion.

Step 1. You first need to get your fraction in a form that you can apply this technique. There are two things are important. First, the fraction needs to be in the simplest form possible with a factorable polynomial in the denominator. The idea of simplest form possible is something you develop over time by working practice problems.
Second, after you satisfy the first requirement, if you have a polynomial in the numerator as well, the highest power in the numerator must be less than the highest power in the denominator. If it isn't, you need to do polynomial long division to get the correct form.
Most of the time, the problem will be given to you in the correct form. But just be aware of these requirements. [ This example is already in the correct form. See the practice problems for problems where you need to do more work. ]

Step 2. Completely factor the denominator using whatever techniques you have.
\(x^2+3x+2 = (x+1)(x+2)\)

Step 3. Write the original fraction as separate fractions, with the denominators as the terms you found in step 2 and with numerators with unknown constants.
\(\displaystyle{ \frac{1}{x^2+3x+2} = }\) \(\displaystyle{ \frac{1}{(x+1)(x+2)} = }\) \(\displaystyle{ \frac{A}{x+1} + \frac{B}{x+2} }\)

Step 4. Multiply both sides of the equation by the factored denominator and cancel like terms.
\(\displaystyle{ \frac{1}{(x+1)(x+2)} [(x+1)(x+2)] }\) \(\displaystyle{ = \left[ \frac{A}{x+1} + \frac{B}{x+2} \right] [(x+1)(x+2)] }\) \(\displaystyle{ 1 = \frac{A}{x+1} [(x+1)(x+2)] + }\) \(\displaystyle{ \frac{B}{x+2} [(x+1)(x+2)] }\) \(\displaystyle{ 1 = A(x+2) + B(x+1) }\)
Notice what happened here. On the left side, the denominator completely canceled. On the right, the \(x+1\) canceled in the first fraction and the \(x+2\) term canceled in the second fraction.

Step 5. Find the value of constants in the numerators.
The equation we have so far must hold for all values of \(x\). So we can choose whatever values of \(x\) we want to solve for \(A\) and \(B\). We suggest you choose values so that one of the terms cancels, for example, in this problem, we would choose \(x=-1\) ( which cancels the term for \(B\)) and then \(x=-2\). Let's see how this works.
Using the last equation above, we have \(1 = A(x+2) + B(x+1) \)

\(x=-1 \to\) \(1 = A(-1+2) + B(-1+1) \to\) \(1 = A(1) + B(0) \to\) \(1 = A \)

\(x = -2 \to\) \( 1 = A(-2+2) + B(-2+1) \to\) \( 1 = A(0) + B(-1) \to\) \(-1 = B \)

Step 6. Final Answer: Clearly show your resulting equations.
\(\displaystyle{ \frac{1}{x^2+3x+2} = \frac{1}{x+1} + \frac{-1}{x+2} }\)

Complications

Okay, so now you are thinking, is that all there is to it? Well, actually, no. We chose a pretty easy problem to show you the basic steps and, in general, those steps do not change from problem to problem. The complication comes in when setting up the original set of fractions, after factoring ( step 3 ). In the above example, we had very simple terms with the highest power of one in both cases. This told us that we just needed constants ( \(A\) and \(B\) ) in the numerators. This is not always the case. The numerators with variables can be more complicated but not overly so. Let's look at each case here. Once you get the pattern down and the equations set up, all the other steps are pretty much the same.
Keep in mind that the terms we are looking at must be completely factored (in the real number system) ( as shown in step 2 above ).

Linear Factors ( Single and Repeating )

When your factor is a linear factor, i.e. \(ax+b\), your numerator is always a constant. A linear factor means the highest power of the variable (\(x\) in this case ) is one and \(a \neq 0 \). Notice, we do not require that \(b\) be non-zero. So the term \(ax\) also falls in this category. The example above had two linear factors, \((x+1)\) and \((x+2)\). So we set up the partial fractions as
\(\displaystyle{ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} }\)

Repeated Factors
What happens when we have a repeated linear factor? A repeating factor means that we have something like \((ax+b)^2\). This case has to be handled differently. We still use constants in the numerator ( since we are still working with linear factors ), but we need extra fractions. Let's set up an example to see what we need to do.

Linear Factors Example
\(\displaystyle{ \frac{1}{x(x+1)^2} = \frac{A_1}{x} + \frac{A_2}{x+1} + \frac{A_3}{(x+1)^2} }\)
Notice that we still use constants in the numerator because we have only linear factors ( one of which is repeated ) in the denominator. Then, we need two terms for the repeated term, one with power one, the second with power two.

In general, it looks like this for linear factors.

factor in the denominator

partial fraction terms

\(ax+b\)

\(\displaystyle{ \frac{A}{ax+b} }\)

\((ax+b)^2\)

\(\displaystyle{ \frac{A_1}{ax+b} +}\) \(\displaystyle{\frac{A_2}{(ax+b)^2} }\)

\((ax+b)^3\)

\(\displaystyle{ \frac{A_1}{ax+b} + }\) \(\displaystyle{\frac{A_2}{(ax+b)^2} + }\) \(\displaystyle{\frac{A_3}{(ax+b)^3} }\)

Do you see the pattern? For a denominator with the term \( (ax+b)^k\), we would have the factors

\(\displaystyle{ \frac{A_1}{ax+b} + }\) \(\displaystyle{ \frac{A_2}{(ax+b)^2} + }\) \(\displaystyle{ \frac{A_3}{(ax+b)^3} + . . . + }\) \(\displaystyle{ \frac{A_k}{(ax+b)^k} }\)

Once these expansions are set up, the steps to find the constants are the same as shown in step 5 above.

Quadratic Factors ( Single and Repeating )

Terms with quadratic factors are of the form \(ax^2+bx+c\) where the highest power on the variable (\(x\) in this case ) is two and \(a \neq 0\). We do not have the same requirement on \(b\) and \(c\), so \(ax^2\) and \(ax^2+c\) are also considered quadratic factors. However, the term \(ax^2+bx\) is NOT considered a quadratic factor, since it can be factored into two simple factors, i.e. \(ax^2+bx = x(ax+b)\). In this case, we follow the techniques associated with simple factors.

When we have a quadratic factor, the numerator must be of the form \(Ax+B\). Notice that we now have an \(x\) in the numerator, not just constants. Also, notice that the highest power of \(x\) is one less than the highest power in the denominator.

Quadratic Factors Example
\(\displaystyle{ \frac{1}{x(x^2+3)} = \frac{A}{x} + \frac{Bx+C}{x^2+3} }\)

Repeated Factors
If you understand how repeated factors work for simple terms discussed in the previous panel, you should be able to anticipate how repeated factors work for quadratic factors.

factor in the denominator

partial fraction terms

\(ax^2+bx+c\)

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} }\)

\((ax^2+bx+c)^2\)

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} +}\) \(\displaystyle{\frac{A_2x+B_2}{(ax^2+bx+c)^2} }\)

\((ax^2+bx+c)^3\)

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} +}\) \(\displaystyle{\frac{A_2x+B_2}{(ax^2+bx+c)^2} +}\) \(\displaystyle{\frac{A_3x+B_3}{(ax^2+bx+c)^3} }\)

Do you see the pattern? For a denominator with the term \( (ax^2+bx+c)^k\), we would have the factors

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + }\) \(\displaystyle{ \frac{A_2x+B_2}{(ax^2+bx+c)^2} + }\) \(\displaystyle{ \frac{A_3x+B_3}{(ax^2+bx+c)^3} + . . . + }\) \(\displaystyle{ \frac{A_kx+B_k}{(ax^2+bx+c)^k} }\)

Once these expansions are set up, the steps to find the constants are the same.

Alternate Technique To Find Constants

The technique above (shown in step 5) works in most cases. However, sometimes the equations can get quite messy. (Not that they can't get messy with this technique too but sometimes this is less complicated.)
Note: As always, check with your instructor to see what they expect. Sometimes, they may be emphasizing something by requiring a certain technique. Their requirements come before those found on this site. But mostly, trust them. They may know what they are doing.

One reservation we have with the technique shown above that you may have thought about is the comment that the equation must hold for all values of \(x\). However, if you look closely, the original fraction is not defined at the points that we are sometimes plugging in for \(x\) since the denominator is zero there. (Go back to the example above if this is not clear.) However, plugging in those values works, although why it works is not clear. Therefore, to avoid this situation, we present this technique that, honestly, is clearer and involves less algebra.

You may ask why didn't just show this technique above. Well, we have found that many, if not most, teachers teach that technique. So we presented it up front. In our opinion, this is the preferred technique, so we will use it in most practice problems below.

Alternate Technique

This technique involves matching coefficients of the variables. Let's use the example above to see how it works. We start out with \(1 = A(x+2) + B(x+1) \), which is the result from step 4. The variable involved is \(x\) and the constants we are trying to find are \(A\) and \(B\). First, we multiply out the right side and combine common terms depending on the variable \(x\).
\( \begin{array}{rcl} 1 & = & A(x+2) + B(x+1) \\ & = & Ax + 2A + Bx + B \\ & = & Ax + Bx + 2A + B \\ & = & (A+B)x + (2A+B) \end{array} \)
Okay, so, if we write out explicitly the terms on the left, so that we can seem them clearly, we have the equation

\( 0x+1 = (A+B)x + (2A+B)\)

Now, we equate the cofficients of the terms to get the two equations

\( 0 = A+B \) and \(1 = 2A+B \)

Now, we have several ways we can solve these, including substitution, matrices and adding/subtracting the equations. We will use the last technique and subtract the first equation from the second.
\( \begin{array}{rcrcl} 2A & + & B & = & 1 \\ -(~~A & + & B & = & 0~~) \\ -A & - & B & = & 0 \\ ---&---&---&---&--- \\ A & & & = & 1 \end{array} \)

Substituting \(A=1\) into the equation \(A+B=0\) gives us \(B=-1\).
These are the same values we obtained from the above technique.

Using Partial Fractions In Integration

Okay, so now that you are comfortable with partial fraction expansion, how can you use it to evaluate integrals. You will need to be able to evaluate integrals with polynomials in the denominator. The idea is to factor the denominator, expand into partial fractions and then evaluate each integral individually, almost always using integration by substitution.

When you have simple factors like \(\displaystyle{ \int{ \frac{1}{x+a} } }\), the substitution is pretty obvious. You let \(u=x+a\) and evaluate, usually ending up with natural logarithms in your answer.

Okay, let's watch a video clip to see how this works. In this video, she explains the technique while going through a specific example. If you don't understand all the steps toward the end, don't worry about it. Specifically, there is an integration that ends up with an inverse trig function. If you haven't had that material yet, you can still get a lot out of this video.

Krista King Math - Solving Integrals - Partial Fractions Clip [3min-3secs]

video by Krista King Math

Rather than continuing to explain this technique in general, it is better to look at an example. Try to work this example on your own before looking at the solution.

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\).

\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } = \ln \abs{\frac{x+1}{x+2}} + C }\)

Problem Statement: Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\).
Solution: Notice we can't use substitution since, if we let \(u=x^2+3x+2\), then \(du=(2x+3)dx\) and we don't have \(2x+3\) in the integrand (and there is no way to get it). So we use the result above (in the steps panel) to expand the integrand and the result is
\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } = }\) \(\displaystyle{ \int{ \frac{1}{x+1} + \frac{-1}{x+2} dx } = }\) \(\displaystyle{ \int{ \frac{1}{x+1} dx } + \int{ \frac{-1}{x+2} dx } = }\) \(\displaystyle{ \ln \abs{x+1} - \ln \abs{x+2} + C = }\) \(\displaystyle{ \ln \abs{\frac{x+1}{x+2}} + C }\)
In this work, we used integration by substitution but we didn't write out the steps. Here are the details for one of the integrals. You can figure out the other one easily after understanding this one.
\(\displaystyle{ \int{ \frac{dx}{x+1} } }\)
Substitution, let \(u = x+1 ~~~ \to ~~~ du = dx\)
\(\displaystyle{ \int{ \frac{dx}{x+1} } = }\) \(\displaystyle{ \int{ \frac{1}{u}du} = }\) \(\displaystyle{ \ln \abs{u} + c_1 = }\) \(\displaystyle{ \ln \abs{x+1} + c_1 }\)
Notice in the above solution that, once we expanded the integrand using partial fractions, we had two fractions that we could then use integration by substitution to solve. THAT'S the point of integration using partial fractions.

Final Answer

\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } = \ln \abs{\frac{x+1}{x+2}} + C }\)

close solution

Things To Watch For

1. Before actually starting to do partial fraction expansion, take a quick look at your integrand and make sure that the highest power in the denominator is larger than the highest power in the numerator. If it isn't, you need to do long division of polynomials or synthetic division to reduce the power in the numerator. This will often make the integral much easier to evaluate before doing any partial fraction expansion.
2. Some problems that do not look like they are candidates for partial fractions may be able to be converted to partial fraction form using substitution. You will often see them when you have \(e^x\) terms. See the practice problems for examples of this.

Practice

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.

Integration Using Partial Fractions - Practice Problems Conversion

[A01-1026] - [A02-1028] - [A03-1029] - [A04-1031] - [A05-1036] - [A06-1038] - [A07-1041] - [A08-1847]

[B01-493] - [B02-1027] - [B03-1030] - [B04-1033] - [B05-1034] - [B06-1037] - [B07-1039] - [B08-1040] - [B09-1035]

[B10-1042] - [B11-1043] - [B12-1032] - [B13-1823]

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

GOT IT. THANKS!

Instructions - Evaluate the following integrals using partial fractions. Give your answers in exact simplified terms.

Basic Problems

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

Problem Statement

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

Final Answer

\(\displaystyle{ \int{ \frac{du}{u^2-1}} = \frac{1}{2} \ln\abs{\frac{u-1}{u+1}} + C }\)

Problem Statement

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

Solution

First, we rewrite the integrand using partial fraction expansion.

\(\displaystyle{ \frac{1}{u^2-1} = \frac{A}{u-1} + \frac{B}{u+1} }\)

Now solve for A and B.

\( 1 = A(u+1) + B(u-1) \)

\( u = 1 ~~~ 1 = A(2) ~~~ \to ~~~ A = 1/2 \)

\( u = -1 ~~~ 1 = B(-2) ~~~ \to ~~~ B = -1/2 \)

Rewrite the integral and evaluate.

\(\displaystyle{ \int{\frac{du}{u^2-1}} = \int{ \frac{1/2}{u-1} + \frac{-1/2}{u+1} du} }\)

\(\displaystyle{ \frac{1}{2} \ln \abs{u-1} - \frac{1}{2} \ln\abs{u+1} + C }\)

\(\displaystyle{ \frac{1}{2} \ln \abs{\frac{u-1}{u+1}}+ C }\)

Final Answer

\(\displaystyle{ \int{ \frac{du}{u^2-1}} = \frac{1}{2} \ln\abs{\frac{u-1}{u+1}} + C }\)

close solution

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

Problem Statement

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

Solution

1028 solution video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \int{ \frac{(2x^2+2x+6)~dx}{(x-1)(x+1)(x-2)} } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{(2x^2+2x+6)~dx}{(x-1)(x+1)(x-2)} } }\)

Solution

1029 solution video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\)

Final Answer

\(-\ln|x+1|+2\ln|x-3|+C\)

Problem Statement

\(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\)

Solution

Here is the link to the blog page for this problem with the written out solution. [link]

2317 solution video

Final Answer

\(-\ln|x+1|+2\ln|x-3|+C\)

close solution

\(\displaystyle{ \int{ \frac{2x-1}{x^2+5x+6} dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{2x-1}{x^2+5x+6} dx } }\)

Solution

1031 solution video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \int{ \frac{5}{(2x+1)(x-2)} dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{5}{(2x+1)(x-2)} dx } }\)

Solution

1036 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1038 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1041 solution video

video by Krista King Math

close solution

\(\displaystyle{ \int{ \frac{5x-4}{(x+1)(x-2)^2} dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{5x-4}{(x+1)(x-2)^2} dx } }\)

Solution

1847 solution video

video by Dr Chris Tisdell

close solution

Intermediate Problems

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

Problem Statement

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

Hint

Let \(u=e^x\) and convert the integral in terms of u. This yields an integral where you can use partial fractions.

Problem Statement

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

Hint

Let \(u=e^x\) and convert the integral in terms of u. This yields an integral where you can use partial fractions.

Solution

493 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int{ \frac{dx}{x(x+1)^3} } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{dx}{x(x+1)^3} } }\)

Solution

1027 solution video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \int{ \frac{8x+9}{(x-2)(x+3)^2} dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{8x+9}{(x-2)(x+3)^2} dx } }\)

Solution

1030 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

1033 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1034 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int{ \frac{4x^2-3x-4}{x^3+x^2-2x} dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{4x^2-3x-4}{x^3+x^2-2x} dx } }\)

Solution

1037 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1039 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1040 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1035 solution video

video by PatrickJMT

close solution

\(\displaystyle{ \int{ \frac{x^3-2x^2+x+1}{x^4+5x^2+4} dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{x^3-2x^2+x+1}{x^4+5x^2+4} dx } }\)

Solution

1042 solution video

video by Krista King Math

close solution

\(\displaystyle{ \int_{1}^{2}{ \frac{x^4+1}{x(x^2+1)^2} dx } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{2}{ \frac{x^4+1}{x(x^2+1)^2} dx } }\)

Solution

1043 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

In the video he does not show the details for the partial fraction expansion.
\(\displaystyle{ \frac{x+6}{x(x^2+2x+3)} = \frac{A}{x}+\frac{Bx+C}{x^2+2x+3} }\)
\(x+6=A(x^2+2x+3)+x(Bx+C)\)

\(x=0\)

\(6=A(3)+0 \to A=2\)

\(x=1\)

\(7=2(1+2+3)+1(B+C) \to B+C=-5\)

\(x=-1\)

\(5=2(1-2+3)-(-B+C) \to B-C=1\)

Adding \(B+C=-5\) to \(B-C=1\) gives us \(2B=-5 \to B=-2\), and since \(B-C=1 \to -2-C=1 \to C=-3\).
This problem also requires you to know how to get an inverse trig function out of an integral.

1032 solution video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \int{ \frac{4x}{x^3+x^2+x+1} ~dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{4x}{x^3+x^2+x+1} ~dx } }\)

Solution

1823 solution video

video by Krista King Math

close solution

\(\displaystyle{ \int{ \frac{x~dx}{(a+bx)(c+fx)} } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{x~dx}{(a+bx)(c+fx)} } }\)

Solution

2211 solution video

video by Michel vanBiezen

close solution

Advanced Problems

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

Problem Statement

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

Solution

2214 solution video

video by Michel vanBiezen

close solution
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