Integration with partial fractions
It is highly recommended that you study the partial fractions page before reading this page. Throughout this page we learn how to find various integrals when the integrands are polynomial fractions. As always be sure to check which format of partial fractions is correct and be sure to check if the fraction is improper first.
Integrating Partial Fractions with Linear Factors
To integrate a fraction where the denominator of the integral can be factorised we must first split the integral into partial fractions. For example, suppose we wish to find 
. We notice that we can factorise the integrand so we separate 
into partial fractions. We must factorise the denominator to identify the linear factors first: 
. It follows that we must find 
and 
such that 
. Hence, 
. That is, 
and 
. Solving gives 
and 
. So, 
. The integral becomes:
Click here to see how to integrate fractions with a linear factor using integration by substitution (see Example 1 after clicking here).
Repeated Factors
To integrate a fraction when one of the factors in the denominator is repeated, we use the usual partial fraction rules. For example, find 
. So, to separate 
into partial fractions we begin by factorising the denominator. Factoring out the 
first, then factorising the quadratic in the brackets gives: 
. Since 
is a repeated factor, we find , and 
such that: 
. We write this as a single fraction and compare the numerator with that of the original fraction: 
. It follows that 
. Equating the coefficients gives 
, 
and 
. Hence, , 
and 
giving 
. The integral becomes:
Click here to how to use integration by substitution (see reverse chain rule). See Example 1 on this page for an example of definite integration with partial fractions.
Integrating with Improper Fractions
When integrating an improper fraction, we must use a specific format when finding partial fractions. For example, find 
. Since the fraction is improper, we express 
in the form 
where 
, 
and 
are constants to be found. The given form is used because, even though there is a 
on the top and an 
on the bottom, the order (highest power) is the same and so the fraction is improper. Adding the fractions by finding a common denominator gives 
. Equating coefficients gives 
, 
and 
. It follows that , 
and 
and so 
. Hence, the integral becomes:
Click here to how to use integration by substitution (see reverse chain rule). See Example 2 on this page for another example of integration of an improper fraction using partial fractions.
Examples
Given that 
, find the exact value of 
.
Solution:
We can write the numerator of the fraction as 
. Since the factor of is repeated, we try partial fractions of the form 
. It follows that 
. Equating coefficients gives 
, 
and 
. It follows that 
, 
, 
. The integral becomes:
![\begin{array}{lll}\int_1^2\frac{1}{x^2(2x-3)}dx&=&\int_1^2\left(-\frac{2}{9x}-\frac{1}{3x^2}+\frac{4}{9(2x-3)}\right)dx\\&=&\left[-\frac{2}{9}\ln\vert x\vert +\frac{1}{3x}+\frac{2}{9}\ln\vert 2x-3\vert\right]_1^2\\&=&\left(-\frac{2}{9}\ln(2) +\frac{1}{6}+0\right)-\left(0+\frac{1}{3}+0\right)\\&=&-\frac{2}{9}\ln(2)-\frac{1}{6}\end{array}](https://studywell.com/wp-content/ql-cache/quicklatex.com-df56d393e05ef131f5fa4ee0152ed888_l3.png "Rendered by QuickLaTeX.com")
Find the integral of 
.
Solution:
The numerator of this fraction has the same order as the denominator and so this fraction is improper. The denominator is also a difference of squares that we can factorise to 
. Hence, we try partial fractions of the form 
. It follows that 
. Equating coefficients in the usual way we obtain 
, 
and 
. The first equation gives 
and the other two then simplify to 
and 
. It follows that , and 
giving 
. The integral becomes:
