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Integration by Partial Fractions

April 5th, 2009 |

Author: Merrin

Differentiation and integration of the elementary functions. are concerns of introductory calculus. One class of elementary functions are the rational functions which are formed as the ratio of two polynomials. At first sight, integrating rational functions can be quite challenging, but by decomposing the functions in a particular way called a partial fractions expansion we can go pretty far in integrating some basic rational functions. There are a lot of algebra techniques used in solving integrals of rational functions such as factoring, polynomial division, solving systems of equations, and completing the square. Review material for how to get the partial fractions expansion is in Chapter 1. I have separated out most of the algebra so we can focus on the calculus of how to integrate the pieces of the partial fractions expansion.

The general problem of integrating rational functions can be expressed a

$\displaystyle \int \cfrac{P(x)dx}{Q(x)}$

In the algorithm for integrating rational functions, the first step is to calculate the quotient q(x) and remainder r(x). The quotient is a simple polynomial and the remainder is another rational function where the degree of the numerator is less than the degree of the denominator

$\cfrac{P(x)}{Q(x)} =q(x)+r(x)\quad \quad \quad \quad \, r(x)=\cfrac{N(x)}{Q(x)}$

Integrating the quotient is just a straightforward integration of a polynomial

$\displaystyle \int \sum _{k=1}^{n}a_{k} x^{k} dx=\sum _{k=1}^{n}\cfrac{a_{k} x^{k+1} }{k+1}$

The remainder can be expressed as a partial fractions expansion then integrated or may already be simple enough to integrate.

Example 1. Find the quotient and the remainder then solve the integral.

$\displaystyle \int \cfrac{x^{3} }{1+x} dx$

Solution 1. Polynomial division works, but here is another method to divide simple expressions.

${\cfrac{x^{3} }{1+x} =\cfrac{(x+1-1)^{3} }{1+x} =\cfrac{(u-1)^{3} }{u} =\cfrac{u^{3} -3u^{2} +3u-1}{u} } \\ {\cfrac{x^{3} }{1+x} =(1+x)^{2} -3(1+x)+3-\cfrac{1}{1+x} } \\ {\cfrac{x^{3} }{1+x} =x^{2} -x+1-\cfrac{1}{1+x} }$

We found the quotient by multiplying out all the polynomials and collecting like terms. Integrate the remainder as a logarithm.

$\displaystyle \int \cfrac{x^{3} dx}{1+x} =\displaystyle \int (x^{2} -x+1-\cfrac{1}{1+x} )dx=\cfrac{x^{3} }{3} -\cfrac{x^{2} }{2} +x-\ln \left|1+x\right|+C$

For more complicated remainders that have multiple factors we have to use the partial fractions theory. If the denominator can be factored completely into linear and quadratic factors things work out very well.

Theorem 1. Form of the partial fractions expansion when the degree of P(x) is less than the degree of Q(x) which has been completely factored into linear and quadratic terms.

${Q(x)=(x-r_{1} )^{j_{1} } ...(x-r_{m} )^{j_{m} } ((x-a_{1} )^{2} +b_{1}^{2} )^{k_{1} } ...((x-a_{n} )^{2} +b_{n}^{2} )^{k_{n} } } \\ {r(x)=\cfrac{P(x)}{Q(x)} =\sum _{i=1}^{m}\sum _{r=1}^{j_{i} }\cfrac{A_{ir} }{(x-r_{i} )^{r} } +\sum _{i=1}^{n}\sum _{r=1}^{k_{i} }\cfrac{B_{ir} x+C_{ir} }{((x-a_{i} )^{2} +b_{i}^{2} )^{r} } }$

This equation basically states that all linear terms in the denominator of the remainder which repeat j times can be written out in a form as

$\cfrac{1}{(x-a)^{j} } \to \cfrac{A_{1} }{(x-a)} + \cfrac{A_{2} }{(x-a)^2}+\cdots+\cfrac{A_{k} }{(x-a)^{j}}$

Quadratic terms in the remainder which repeat k times are written out in a form as

$\cfrac{1}{((x-a)^{2} +b^{2} )^{k } } \to \cfrac{A_{1} +B_{1} x}{(x-a)^{2} +b^{2} } + \cfrac{A_{1} +B_{1} x}{((x-a)^{2} +b^{2})^2 }+...+\cfrac{A_{k} +B_{k} x}{((x-a)^{2} +b^{2} )^{k} }$

I used the form with the completion of the square because it is most handy for u substitutions as well as integrating pieces like arctangents.

Example 2. Write out the form of the partial fractions expansion for some representative examples but do not solve for the coefficients.

Solution 2. Apply the partial fraction expansion theorem equation.

$r_{1} (x)=\cfrac{x^{4} +2x+1}{x^{5} } =\cfrac{A}{x} +\cfrac{B}{x^{2} } +\cfrac{C}{x^{3} } +\cfrac{D}{x^{4} } +\cfrac{E}{x^{5} }$

$r_{2} (x)=\cfrac{x^{3} +2x+1}{(x^{2} +3x+6)^{2} } =\cfrac{Ax+B}{(x^{2} +3x+6)} +\cfrac{Cx+D}{(x^{2} +3x+6)^{2} }$

$r_{3} (x)=\cfrac{x^{5} +x+1}{x^{3} (x+1)(x+2)} =\cfrac{A}{x} +\cfrac{B}{x^{2} } +\cfrac{C}{x^{3} } +\cfrac{D}{x+1} +\cfrac{E}{x+2}$

$r_{4} (x)=\cfrac{x^{8} +5x^{4} -2x+1}{x(x^{2} +4x+7)^{3} } =\cfrac{A}{x} +\cfrac{Bx+C}{(x^{2} +4x+7)} +\cfrac{Dx+E}{(x^{2} +4x+7)^{2} } +\cfrac{Fx+G}{(x^{2} +4x+7)^{3} }$

$\, r_{5} (x)=\cfrac{x^{5} -x^{4} +x^{3} +2}{x^{2} (x^{2} +1)^{2} } =\cfrac{A}{x} +\cfrac{B}{x^{2} } +\cfrac{Cx+D}{(x^{2} +1)} +\cfrac{Ex+F}{(x^{2} +1)^{2} }$

Anyone should be able to get the hang of writing out the partial fraction expansion by following these examples. Solving for the coefficients is a matter of algebra.

Example 3. Show how to integrate the different possible pieces in the partial fractions expansion in general.

$\displaystyle \int \cfrac{dx}{(x-a)^{k} } \qquad \displaystyle \int \cfrac{(Ax+B)dx}{(x^{2} +2bx+c^{2} )^{k} }$

Solution 3. The linear pieces are simpler to integrate than the quadratic terms so we will start with them. The forms derived from the linear pieces look like

$(x-a)^{-1} \quad \quad (x-a)^{-2} \qquad (x-a)^{-k}$

$\int \cfrac{1}{x-a} dx=\ln \left|x-a\right| +C} \\ {\displaystyle \int \cfrac{1}{(x-a)^{2} } =-\cfrac{1}{x-a} +C} \\ {\displaystyle \int \cfrac{1}{(x-a)^{k} } =\cfrac{1}{1-k} (x-a)^{1-k} +C$

When k is one the integral is a logarithm otherwise the power rule for integration gives the answer..

Now we move on to the quadratic factors. Quadratic factors in the denominator can be turned into the following form by completing the square.

$\cfrac{Ax+B}{(x-a)^{2} +b^{2} }$

Integration of a quadratic term is done in terms of arctangent and a logarithm.

$\int \cfrac{Ax+B}{(x-a)^{2} +b^{2} } dx } = {\displaystyle \int \cfrac{A(x-a)+B+aA}{(x-a)^{2} +b^{2} } dx$

$\displaystyle \int \cfrac{Au}{u^2+b}=\cfrac{A}{2}\ln\left|(x-a)^2+b\right|+C$

$\displaystyle \int \cfrac{B+aA}{u^2+b^2}=(B+aA)\cfrac{\text{arctan}\,\,((x-a)/b)}{b}+C$

For quadratic factors in the denominator to powers higher than one, trigonometric substitutions are useful to evaluate the second part of the integral

$\int \cfrac{Ax+B}{((x-a)^{2} +b^{2} )^{k} } dx } = {\displaystyle \int \cfrac{A(x-a)+B+aA}{((x-a)^{2} +b^{2} )^{k} } dx=I_{1} +I_{2}$

$I_1=\displaystyle \int \cfrac{A(x-a)dx}{((x-a)^2+b^2)^{k}}=\cfrac{A/2}{1-k}((x-a)^2+b^2)^{1-k}+C$

${I_{2} }={\displaystyle \int \cfrac{(B+aA)dx}{((x-a)^{2} +b^{2} )^{k} } =}$

$u={x-a \quad\ du=dx\quad } \\ u = {b\text{tan}\,\, \theta \quad du=b\text{sec}\,\, ^{2} \theta d\theta } \\ I_2=\displaystyle \int \cfrac{(B+aA)}{(u^2+b^2)^k}du=\displaystyle \int \cfrac{(B+aA)\text{cos}\,\,^{2k-2} \theta}{b^{2k}}d\theta$

Trigonometric integrals such as this can be done for k > 2 but usually the partial fractions examples are not this nasty. Review the section on trigonometric integrals section.

There is a lot of information in this section. The main ideas are summarized here in this next example.

Example 4. Describe the full algorithm for solving integrals of rational functions by partial fractions.

Solution 4. The summary of the steps is as follows

1. Write the rational function as the ratio of a numerator P(x) and a denominator Q(x)

2. If the degree of P(x) is less then Q(x), continue on to the partial fraction expansion step otherwise divide P(x) by Q(x) by polynomial long division and find the remainder r(x) and the quotient q(x)

3. The quotient is just a polynomial and can be easily integrated by the power rule for integration.

4. Expand the remainder as a partial fraction in the proper form.

5. Solve for the coefficients of the terms in the partial fractions expansion, usually the algebra simplifies by plugging in at the roots.

6. Integrate each term in the partial fractions expansion. There are only a finite set of possible integrals which we know how to solve ahead of time.

Example 5. Find

$\displaystyle \int \cfrac{x^{2} +2x+1}{x(x+1)(x-1)} dx$

Solution 5. The degree of the numerator is less than the denominator so the quotient is zero. The remainder can be expressed as a partial fractions expansion.

$r(x)=\cfrac{x^{2} +2x+1}{x(x+1)(x-1)} =\cfrac{A}{x} +\cfrac{B}{x+1} +\cfrac{C}{x-1}$

To solve for the coefficients combine all the fractions on the right and substitute values at the points where the factor equals zero to simplify the algebra.

$x^{2} +2x+1=A(x+1)(x-1)+Bx(x-1)+Cx(x+1)$

Put x=0, -1, and 1

$x=0:\qquad 1=A(1)(-1)+0+0\to A=1$

$x=-1:\qquad 1-2+1=A(0)(-2)+B(-1)(-2)+C(-1)(0)\to B=0$

$x=1:\qquad 1+2+1=A(2)(0)+B(1)(0)+C(1)(2)\to C=1/2$

Write out the form of the remainder function then integrate it.

$r(x)=\cfrac{1}{x} +\cfrac{1}{2(x-1)}$

$\int \cfrac{x^{3} +2x+1}{x^{2} (x+1)(x-1)} dx }={\displaystyle \int r(x)dx=\displaystyle \int \cfrac{dx}{x} +\displaystyle \int \cfrac{dx}{2(x-1)}$

$r(x)=\ln\left| x \right|+\cfrac{1}{2}\ln \left| x-1 \right|+C$

Example 6. Find the integral

$\displaystyle \int \cfrac{3x^{3} +2}{x(x^{2} +1)^{2} } dx$

Solution 6. This expression contains linear as well as quadratic terms. We don’t need to find the integral of the quotient because there is not one. The degree of the numerator is lower than the denominator. We write the remainder out as a partial fraction expansion.

$r(x)=\cfrac{A}{x} +\cfrac{Bx+C}{(x^{2} +1)} +\cfrac{Dx+E}{(x^{2} +1)^{2} }$

There is a straightforward algebra step here to find the coefficients which I will omit

$r(x)=\cfrac{2}{x} -\cfrac{2x}{1+x^{2} } +\cfrac{x}{(1+x^{2} )^{2} }$

Integrating the first two pieces we have

$\displaystyle \int \cfrac{2dx}{x} =2\ln \left|x\right|\quad \quad -2\displaystyle \int \cfrac{xdx}{1+x^{2} } =-\ln \left|1+x^{2} \right|$

Integrating the last piece we have

$\displaystyle \int \cfrac{xdx}{(1+x^{2} )^{2} } =\cfrac{1}{2} \displaystyle \int \cfrac{du}{u^{2} } =-\cfrac{1}{2} u^{-1} =-\cfrac{1/2}{(1+x^{2} )}$

Combine all the pieces and we are done.

$\displaystyle \int \cfrac{3x^{3} +2}{x(x^{2} +1)^{2} } dx =-\cfrac{1/2}{(1+x^{2} )} +\ln \left(\cfrac{x^{2}}{1+x^{2} }\right) +C$

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2 Responses to “Integration by Partial Fractions”

Ian George says:

May 8, 2009 at 6:59 pm

I hada question about the power rule of integration. I wrote a different equation in terms of “x” to the “n” to integrate fractions and was wondering if someone has alerady done this and thought that people from a calculus website might know. please email me back letting me know and i will contact you with my equation.

Internet Banking says:

February 4, 2010 at 9:08 am

Glad to see that this site works well on my iPhone , everything I want to do is functional. Thanks for keeping it up to date with the latest.

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Integration by Partial Fractions

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