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17Calculus - Integration Using Partial Fractions

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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 thing you need to do is to go to the precalculus partial fraction expansion page and review the partial fraction expansion technique and work a few practice problems. This will get you ready to evaluate integrals using partial fractions.

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

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

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

Solution

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

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

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

Evaluate these integrals using partial fractions. Give your answers in exact simplified terms.

Basic

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{du}{u^2-1} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Final Answer

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{du}{u^2-1} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

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.

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

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

\(\displaystyle{ \frac{1}{2} \ln \left| \frac{u-1}{u+1} \right| + C }\)

Final Answer

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

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x+3}{(x+2)^3} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1028 video

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{(2x^2+2x+6)~dx}{(x-1)(x+1)(x-2)} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1029 video

video by Dr Chris Tisdell

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Final Answer

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

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

2317 video

Final Answer

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

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{2x-1}{x^2+5x+6} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1031 video

video by Dr Chris Tisdell

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{5}{(2x+1)(x-2)} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1036 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2+x} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1038 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^3+x} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1041 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{5x-4}{(x+1)(x-2)^2} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1847 video

video by Dr Chris Tisdell

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Intermediate

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{1+e^x}{1-e^x}dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

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

Evaluate \(\displaystyle{ \int{ \frac{1+e^x}{1-e^x}dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

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 video

video by PatrickJMT

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x(x+1)^3} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1027 video

video by Dr Chris Tisdell

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{8x+9}{(x-2)(x+3)^2} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1030 video

video by Dr Chris Tisdell

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{\sqrt{x+1}}{x} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1033 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{1}{x-\sqrt{x+2}} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1034 video

video by PatrickJMT

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{4x^2-3x-4}{x^3+x^2-2x} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1037 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x^2}{(x+2)^3} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1039 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x^2+x+1}{(2x+1)(x^2+1)} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1040 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x^2-x+6}{x^3+3x} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1035 video

video by PatrickJMT

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x^3-2x^2+x+1}{x^4+5x^2+4} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1042 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int_{1}^{2}{ \frac{x^4+1}{x(x^2+1)^2} dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1043 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{\int{\frac{x+6}{x(x^2+2x+3)}dx}}\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

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

\(x=0\)

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

\(x=1\)

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

\(x=-1\)

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

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 video

video by Dr Chris Tisdell

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{4x}{x^3+x^2+x+1} ~dx } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

1823 video

video by Krista King Math

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{x~dx}{(a+bx)(c+fx)} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

2211 video

video by Michel vanBiezen

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Advanced

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

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2(a+bx)^2} } }\) using partial fractions. Give your answer in exact, simplified, factored form.

Solution

2214 video

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Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Evaluate these integrals using partial fractions. Give your answers in exact simplified terms.

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