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 \int\frac{5-3x}{x^2+x-6}dx. We notice that we can factorise the integrand so we separate \frac{5-3x}{x^2+x-6} into partial fractions. We must factorise the denominator to identify the linear factors first: x^2+x-6=(x+3)(x-2). It follows that we must find A and B such that  \frac{5-3x}{x^2+x-6}=\frac{A}{x+3}+\frac{B}{x-2}=\frac{A(x-2)+B(x+3)}{(x+3)(x-2)}. Hence,  5-3x=A(x-2)+B(x+3)=(A+B)x-2A+3B. That is, A+B=-3 and -2A+3B=5. Solving gives A=-\frac{14}{5} and =-\frac{1}{5}. So, \frac{5-3x}{x^2+x-6}=-\frac{\frac{14}{5}}{x+3}-\frac{\frac{1}{5}}{x-2}=-\frac{14}{5(x+3)}-\frac{1}{5(x-2)}. The integral becomes:

\begin{array}{lll}\int\frac{5-3x}{x^2+x-6}dx&=&\int\left(-\frac{14}{5(x+3)}-\frac{1}{5(x-2)}\right)dx \\&=&-\frac{14}{5}\int\frac{1}{(x+3)}dx-\frac{1}{5}\int\frac{1}{(x-2)}dx\\&=&-\frac{14}{5}\ln{\vert x+3\vert}-\frac{1}{5}\ln{\vert x-2\vert}+c\end{array}

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 \int\frac{5x-2}{x^3-2x^2+x}dx. So, to separate \frac{5x-2}{x^3-2x^2+x} into partial fractions we begin by factorising the denominator. Factoring out the x first, then factorising the quadratic in the brackets gives: x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2. Since x-1 is a repeated factor, we find A, B and C such that: \frac{5x-2}{x^3-2x^2+x}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}. We write this as a single fraction and compare the numerator with that of the original fraction: \frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}=\frac{A(x-1)^2}{x(x-1)^2}+\frac{Bx(x-1)}{x(x-1)^2}+\frac{Cx}{x(x-1)^2}=\frac{A(x-1)^2+Bx(x-1)+Cx}{x(x-1)^2}. It follows that A(x-1)^2+Bx(x-1)+Cx=5x-2. Equating the coefficients gives A+B=0, -2A-B+C=5 and A=-2. Hence, A=-2, B=2 and C=3 giving \frac{5x-2}{x^3-2x^2+x}=-\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}. The integral becomes:

\begin{array}{lll}\int\frac{5x-2}{x^3-2x^2+x}dx&=&\int\left(-\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}\right)dx\\&=&-2\ln\vert x\vert+2\ln\vert x-1\vert -\frac{3}{x-1}+c\end{array}

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 \int \frac{3x^2+8x-5}{x^2-6x+9}dx. Since the fraction is improper, we express \frac{3x^2+8x-5}{x^2-6x+9} in the form a+\frac{b}{x-3}+\frac{c}{(x-3)^2} where a, b and c are constants to be found. The given form is used because, even though there is a 3x^2 on the top and an x^2 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 a+\frac{b}{x-3}+\frac{c}{(x-3)^2}=\frac{a(x-3)^2}{(x-3)^2}+\frac{b(x-3)}{(x-3)^2}+\frac{c}{(x-3)^2}=\frac{a(x-3)^2+b(x-3)+c}{(x-3)^2}. Equating coefficients gives a=3, -6a+b=8 and 9a-3b+c=-5. It follows that a=3, b=26 and c=46 and so \frac{3x^2+8x-5}{x^2-6x+9}=3+\frac{26}{x-3}+\frac{46}{(x-3)^2}. Hence, the integral becomes:

\begin{array}{lll}\int \frac{3x^2+8x-5}{x^2-6x+9}dx&=&\int\left(3+\frac{26}{x-3}+\frac{46}{(x-3)^2}\right)dx\\&=&3x+26\ln\vert x-3\vert -\frac{46}{x-3}+c\end{array}

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.


Given that f(x)=\frac{1}{2x^2-3x^3}, find the exact value of \int_1^2f(x)dx.


We can write the numerator of the fraction as 2x^2-3x^3=x^2(2x-3). Since the factor of x is repeated, we try partial fractions of the form \frac{A}{x}+\frac{B}{x^2}+\frac{C}{2x-3}. It follows that \frac{Ax(2x-3)+B(2x-3)+Cx^2}{x^2(2x-3)}\equiv \frac{1}{x^2(2x-3)}. Equating coefficients gives 2A+C=0, -3A+2B=0 and -3B=1. It follows that A=-\frac{2}{9}, B=-\frac{1}{3}, C=\frac{4}{9}. 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}

Find the integral of y=\frac{25x^2}{25x^2-16}.


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 (5x-4)(5x+4). Hence, we try partial fractions of the form A+\frac{B}{5x-4}+\frac{C}{5x+4}. It follows that \frac{A(5x-4)(5x+4)+B(5x+4)+C(5x-4)}{(5x-4)(5x+4)}\equiv \frac{25x^2}{(5x-4)(5x+4)}. Equating coefficients in the usual way we obtain 25A=25, 5B+5C=0 and -16A+4B-4C=0. The first equation gives A=1 and the other two then simplify to B+C=0 and B-C=4. It follows that A=1, B=2 and C=-2 giving \frac{25x^2}{25x^2-16}\equiv 1+\frac{2}{5x-4}-\frac{2}{5x+4}. The integral becomes:

\begin{array}{lll}\int \frac{25x^2}{25x^2-16}dx&=&\int\left(1+\frac{2}{5x-4}-\frac{2}{5x+4}\right)dx\\&=&x+\frac{2}{5}\ln\vert 5x-4\vert-\frac{2}{5}\ln\vert 5x+4\vert+c \end{array}