
Consider the integral \[\int \frac{3x^32x^219x7}{x^2x6}\, dx.\] The integrand is an improper rational function. By "long division" of polynomials, we can rewrite the integrand as the sum of a polynomial and a proper rational function "remainder": \[\frac{3x^32x^219x7}{x^2x6}=3x+1+\frac{1}{x^2x6}.\] So \[\int \frac{3x^32x^219x7}{x^2x6}\, dx=\int \left(3x+1+\frac{1}{x^2x6}\right)\, dx.\] This looks much easier to work with! We can integrate $3x+1$ immediately, but what about $\displaystyle \frac{1}{x^2x6}$? Notice that \[\frac{1}{x^2x6}=\frac{1}{(x+2)(x3)}\] which suggests that we try to write $\displaystyle \frac{1}{x^2x6}$ as the sum of two rational functions of the form $\displaystyle\frac{A}{x+2}$ and $\displaystyle\frac{B}{x3}$: \[\frac{1}{x^2x6}=\frac{A}{x+2}+\frac{B}{x3}.\] This is called the Partial Fraction Decomposition for $\displaystyle \frac{1}{x^2x6}$. Our goal now is to determine $A$ and $B$. Multiplying both sides of the equation by $(x+2)(x3)$ to clear the fractions, \[1=A(x3)+B(x+2).\] There are two methods for solving for $A$ and $B$:
So \[\frac{1}{x^2x6}=\frac{\frac{1}{5}}{x+2}\frac{\frac{1}{5}}{x3}.\] Returning to the original integral, \begin{eqnarray*} \int \frac{3x^32x^219x7}{x^2x6}\, dx&=&\int \left(3x+1+\frac{\frac{1}{5}}{x+2}\frac{\frac{1}{5}}{x3}\right)\, dx\\ &=&\frac{3}{2}x^2+x+\frac{1}{5}\ln \left\frac{x+2}{x3}\right+C. \end{eqnarray*} In the next example, we have repeated factors in the denominator, as well as an irreducible quadratic factor.
ExampleWe will evaluate \[\int \frac{x1}{x^2(x^2+x+1)}\, dx.\] The integrand is a proper rational function, which we would like to decompose into proper rational functions of the form \[\frac{A}{x},\quad \frac{B}{x^2},\quad \mathrm{and}\quad \frac{Cx+D}{x^2+x+1}.\] [Notice that we have two factors of $x$ in the denominator of the integrand, leading to terms of the form $\displaystyle\frac{A}{x}$ and $\displaystyle\frac{B}{x^2}$ in the decomposition. The factor $x^2+x+1$ is irreducible and quadratic, so any proper rational function with $x^2+x+1$ as denominator has the form $\displaystyle\frac{Cx+D}{x^2+x+1}$ where $C$ or $D$ may be $0$.] Set \[\frac{x1}{x^2(x^2+x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+x+1}. \] Multiplying through by $x^2(x^2+x+1)$, \[x1=Ax(x^2+x+1)+B(x^2+x+1)+(Cx+D)x^2.\] Since $x^2+x+1$ has no real roots, it is easiest to solve for $A$ and $B$ using Method 1: Collecting like terms on the right, \[x1=(A+C)x^3+(A+B+D)x^2+(A+B)x+B.\] Equating corresponding powers of $x$, \[ \left.\begin{array}{rcr} A+C&=&0\\ A+B+D&=&0\\ A+B&=&1\\ B&=&1 \end{array}\right\}\quad\longrightarrow\quad \begin{array}{l} A=2\\ B=1\\ C=2\\ D=1 \end{array}\quad\longrightarrow\quad \frac{2}{x}\frac{1}{x^2}\frac{2x+1}{x^2+x+1}. \] So \begin{eqnarray*} \frac{x1}{x^2(x^2+x+1)}\, dx&=&\int\left(\frac{2}{x}\frac{1}{x^2}\frac{2x+1}{x^2+x+1}\right)\, dx\\ &=&2\ln x+\frac{1}{x}\ln x^2+x+1+C\\ &=&\frac{1}{x}+\ln\left\frac{x^2}{x^2+x+1}\right+C. \end{eqnarray*}
Key Concepts
Partial Fraction Decomposition of a Rational Function
The partial fraction decomposition is often used to rewrite a complicated rational function integrand as a sum of terms, each of which is straightforward to integrate. 