In this page, we show how to integrate any rational function ( a ratio of polynomials ) by expressing it as a sum of partial fractions.

Let $f(x)$ be a rational function such that

$$f(x)=\frac{P(x)}{Q(x)}=\frac{a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0}{b_m{x}^m+b_{m-1}{x}^{m-1}+\cdots +b_0}$$

where $P(x)$ and $Q(x)$ are polynomials. If $a_n\ne 0$, the degree of P is n and we denote $\text{deg}(P)=n$.

To integrate any rational function, see the following steps.

Example: Evaluate $\int \frac{2x^5-4x^4-5x^3+9x^2}{x^4-2x^3-3x^2}dx$

Let $f(x)=\frac{P(x)}{Q(x)}=\frac{2x^5-4x^4-5x^3+9x^2}{x^4-2x^3-3x^2}$

$$\int \frac{2x^5-4x^4-5x^3+9x^2}{x^4-2x^3-3x^2}dx=\int \frac{x^2(2x^3-4x^2-5x+9)}{x^2(x^2-2x-3)}dx=\int \frac{2x^3-4x^2-5x+9}{x^2-2x-3}dx$$

$$\frac{2x^3-4x^2-5x+9}{x^2-2x-3}=\frac{2x(x^2-2x-3)+(x-9)}{x^2-2x-3}=2x+\frac{x-9}{x^2-2x-3}$$

$$2x+\frac{x-9}{x^2-2x-3}=2x+\frac{x-9}{(x+1)(x-3)}$$

$$2x+\frac{x-9}{(x+1)(x-3)}=2x+\frac{-2}{x+1}+\frac{3}{x-3}$$

Thus

$$\begin{array}{rcl}\int \frac{2x^5-4x^4-5x^3+9x^2}{x^4-2x^3-3x^2}dx& =& \int (2x+\frac{-2}{x+1}+\frac{3}{x-3})dx\\ & =& \int 2xdx-2\int \frac{1}{x+1}dx+3\int \frac{1}{x-3}dx\\ & =& x^2-2\ln |x+1|+3\ln |x-3|+C\\ & =& x^2+\ln \frac{|x-3{|}^3}{(x+1{)}^2}+C\end{array}$$

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Express as a sum of partial fraction

There are theorems in Algebra that grantee, any rational function can be expressed as a sum of partial fractions

We see the details for the four cases occur.

Case 1:$Q(x)=(a_{1}x+b_1)(a_{2}x+b_2)\cdots (a_{m}x+b_m)$

In the case of that $Q(x)$ is a product of distinct linear factors;

$$Q(x)=(a_{1}x+b_1)(a_{2}x+b_2)\cdots (a_{m}x+b_m)$$

where no factor is repeated and no factor is a constant multiple of another, the fraction can be decomposed as below:

$$\frac{R(x)}{Q(x)}=\frac{A_1}{a_{1}x+b_1}+\frac{A_2}{a_{2}x+b_2}+\cdots \frac{A_m}{a_{m}x+b_m}.$$

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Example : $\frac{x+9}{x^2-2x-3}$

Since the two distinct linear factors, the partial fraction decomposition has the form;

$$\frac{x+9}{x^2-2x-3}=\frac{x+9}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}$$

To determine the value of A and B, we multiply both sides of the equation by the product of denominators, $(x+1)(x-3)$, obtaining

$$x+9=A(x-3)+B(x+1)\cdots (\ast )$$

Solving, we get $A=-2$ and $B=3$.

Case 2:$Q(x)=(a_{1}x+b_1{)}^{r_1}(a_{2}x+b_2{)}^{r_2}\cdots (a_{m}x+b_m{)}^{r_m}$

In the case of that $Q(x)$ is a product of linear factors, some of which are repeated;

$$Q(x)=(a_{1}x+b_1{)}^{r_1}(a_{2}x+b_2{)}^{r_2}\cdots (a_{m}x+b_m{)}^{r_m}$$

the fraction can be decomposed as below:

$$\begin{array}{rcl}\frac{R(x)}{Q(x)}& =& \frac{A_{11}}{a_{1}x+b_1}+\frac{A_{12}}{(a_{1}x+b_1{)}^2}+\cdots +\frac{A_{1r_1}}{(a_{1}x+b_1{)}^{r_1}}\\ & & +\cdots +\\ & & \frac{A_{m1}}{a_{m}x+b_m}+\frac{A_{m{r}_2}}{(a_{m}x+b_m{)}^2}+\cdots +\frac{A_{m{r}_m}}{(a_{m}x+b_m{)}^{r_m}}\end{array}.$$

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Example $\frac{2x+3}{x^2(x+1{)}^2}$

$$\frac{2x+3}{x^2(x+1{)}^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{(x+1{)}^2}$$

Multiplying by the least common denominator, $x^2(x+1{)}^2$, we get

$$2x+3=Ax(x+1{)}^2+B(x+1{)}^2+C{x}^2(x+1)+D{x}^2$$

and equating coefficients, we obtain $A=-4,B=3,C=4$ and $D=1$.

Case 3:$Q(x)=\prod _{n=1}^m(a_{n}x+b_n{)}^{r_n}\prod _{\ell =1}^k(A_{\ell }{x}^2+B_{\ell }x+C_{\ell })$

In the case of that $Q(x)$ contains irreducible quadric factors, none of which is repeated (i.e; $Q(x)$ has the factor $A{x}^2+Bx+C$ where $B^2-4AC<0$);

$$\begin{array}{rcl}Q(x)=& & \prod _{n=1}^m(a_{n}x+b_n{)}^{r_n}\prod _{\ell =1}^k(A_{\ell }{x}^2+B_{\ell }x+C_{\ell })\\ =& & (a_{1}x+b_1{)}^{r_1}\cdots (a_{m}x+b_m{)}^{r_m}\\ & & \times (A_1{x}^2+B_{1}x+C_1)\cdots (A_k{x}^2+B_{k}x+C_k)\end{array}$$

the fraction can be decomposed as below:

$$\begin{array}{rcl}\frac{R(x)}{Q(x)}& =& \sum _{n=1}^m\sum _{i=1}^{r_n}\frac{A_{ni}}{(a_{n}x+b_n{)}^i}+\sum _{\ell =1}^k\frac{D_{\ell }x+E_{\ell }}{A_{\ell }{x}^2+B_{\ell }x+C_{\ell }}\end{array}.$$

Example : $\frac{2x+3}{x^2(x^2+1)}$

$$\frac{2x+3}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}$$

Multiplying by the least common denominator, $x^2(x^2+1)$, we get

$$2x+3=Ax(x^2+1)+B(x^2+1)+(Cx+D)x^2$$

and equating coefficients, we obtain $A=2,B=3,C=-2$ and $D=-3$.

Case 4: $Q(x)=\prod _{n=1}^m(a_{n}x+b_n{)}^{r_n}\prod _{\ell =1}^k(A_{\ell }{x}^2+B_{\ell }x+C_{\ell }{)}^{R_{\ell }}$

In the case of that $Q(x)$ contains irreducible quadric factors, some of which are repeated;

$$\begin{array}{rcl}Q(x)=& & \prod _{n=1}^m(a_{n}x+b_n{)}^{r_n}\prod _{\ell =1}^k(A_{\ell }{x}^2+B_{\ell }x+C_{\ell }{)}^{R_{\ell }}\\ =& & (a_{1}x+b_1{)}^{r_1}\cdots (a_{m}x+b_m{)}^{r_m}\\ & & \times (A_1{x}^2+B_{1}x+C_1{)}^{R_1}\cdots (A_k{x}^2+B_{k}x+C_k{)}^{R_k}\end{array}$$

the fraction can be decomposed as below:

$$\begin{array}{rcl}\frac{R(x)}{Q(x)}& =& \sum _{n=1}^m\sum _{i=1}^{r_n}\frac{A_{ni}}{(a_{n}x+b_n{)}^i}+\sum _{\ell =1}^k\sum _{j=1}^{R_{\ell }}\frac{D_{\ell }x+E_{\ell }}{(A_{\ell }{x}^2+B_{\ell }x+C_{\ell }{)}^j}\end{array}.$$

Example : $\frac{2x+3}{x^2(x^2+1{)}^2}$

$$\frac{2x+3}{x^2(x^2+1{)}^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}+\frac{Ex+F}{(x^2+1{)}^2}$$

Multiplying by the least common denominator, $x^2(x^2+1{)}^2$, we get

$$2x+3=Ax(x^2+1{)}^2+B(x^2+1{)}^2+(Cx+D)x^2(x^2+1)+(Ex+F)x^2$$

and equating coefficients, we obtain $A=2,B=3,C=-2,D=-3,E=-2$ and $F=-3$.