17Calculus - Integration Using Partial Fractions
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 |
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Partial Fraction Expansion Precalculus Review |
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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
Using Partial Fractions In Integration |
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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 |
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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 |
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\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } = \ln \abs{\frac{x+1}{x+2}} + C }\) |
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Things To Watch For |
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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
Instructions - Evaluate the following integrals using partial fractions. Give your answers in exact simplified terms.
Basic Problems |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{du}{u^2-1} } }\)
Final Answer |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{du}{u^2-1} } }\)
Solution |
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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 |
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\(\displaystyle{ \int{ \frac{du}{u^2-1}} = \frac{1}{2} \ln\abs{\frac{u-1}{u+1}} + C }\) |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x+3}{(x+2)^3} dx } }\)
Solution |
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1028 video
video by Dr Chris Tisdell |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{(2x^2+2x+6)~dx}{(x-1)(x+1)(x-2)} } }\)
Solution |
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1029 video
video by Dr Chris Tisdell |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\)
Final Answer |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x+5}{x^2-2x-3} ~dx } }\)
Solution |
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Here is the link to the blog page for this problem with the written out solution. [link]
2317 video
video by World Wide Center of Mathematics |
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Final Answer |
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\(-\ln|x+1|+2\ln|x-3|+C\) |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{2x-1}{x^2+5x+6} dx } }\)
Solution |
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1031 video
video by Dr Chris Tisdell |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{5}{(2x+1)(x-2)} dx } }\)
Solution |
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1036 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{dx}{x^2+x} } }\)
Solution |
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1038 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{dx}{x^3+x} } }\)
Solution |
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1041 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{5x-4}{(x+1)(x-2)^2} dx } }\)
Solution |
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1847 video
video by Dr Chris Tisdell |
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Intermediate Problems |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{1+e^x}{1-e^x}dx } }\)
Hint |
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Let \(u=e^x\) and convert the integral in terms of u. This yields an integral where you can use partial fractions.
Problem Statement |
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\(\displaystyle{ \int{ \frac{1+e^x}{1-e^x}dx } }\)
Hint |
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Let \(u=e^x\) and convert the integral in terms of u. This yields an integral where you can use partial fractions.
Solution |
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493 video
video by PatrickJMT |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{dx}{x(x+1)^3} } }\)
Solution |
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1027 video
video by Dr Chris Tisdell |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{8x+9}{(x-2)(x+3)^2} dx } }\)
Solution |
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1030 video
video by Dr Chris Tisdell |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{\sqrt{x+1}}{x} dx } }\)
Solution |
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1033 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{1}{x-\sqrt{x+2}} dx } }\)
Solution |
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1034 video
video by PatrickJMT |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{4x^2-3x-4}{x^3+x^2-2x} dx } }\)
Solution |
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1037 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x^2}{(x+2)^3} dx } }\)
Solution |
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1039 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x^2+x+1}{(2x+1)(x^2+1)} dx } }\)
Solution |
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1040 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x^2-x+6}{x^3+3x} dx } }\)
Solution |
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1035 video
video by PatrickJMT |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x^3-2x^2+x+1}{x^4+5x^2+4} dx } }\)
Solution |
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1042 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int_{1}^{2}{ \frac{x^4+1}{x(x^2+1)^2} dx } }\)
Solution |
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1043 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{\int{\frac{x+6}{x(x^2+2x+3)}dx}}\)
Solution |
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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 video
video by Dr Chris Tisdell |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{4x}{x^3+x^2+x+1} ~dx } }\)
Solution |
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1823 video
video by Krista King Math |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{x~dx}{(a+bx)(c+fx)} } }\)
Solution |
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2211 video
video by Michel vanBiezen |
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Advanced Problems |
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Problem Statement |
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\(\displaystyle{ \int{ \frac{dx}{x^2(a+bx)^2} } }\)
Solution |
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2214 video
video by Michel vanBiezen |
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