Question

Evaluate the following integrals:

$\int \frac{x^2}{\left(x^2+4\right)\left(x^2+9\right)}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2}{\left(x^2+4\right)\left(x^2+9\right)}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{x^2}{\left(x^2+4\right)\left(x^2+9\right)}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+9} \\ \Rightarrow x^2=\left(Ax+B\right)\left(x^2+9\right)+\left(Cx+D\right)\left(x^2+4\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^3,x^2,x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 0=A+C\mathrm{and}1=B+D\mathrm{and}0=9A+4C\mathrm{and}0=9B+4D \\ \mathrm{or}A=0\mathrm{and}B=-\frac{4}{5}\mathrm{and}C=0\mathrm{and}D=\frac{9}{5} \\ \therefore I=\int \left(\frac{{\displaystyle -\frac{4}{5}}}{x^2+4}+\frac{{\displaystyle \frac{9}{5}}}{x^2+9}\right)\mathrm{d}x \\ =-\frac{4}{5}\int \frac{1}{x^2+4}\mathrm{d}x+\frac{9}{5}\int \frac{1}{x^2+9}\mathrm{d}x \\ =-\frac{4}{5}\times \frac{1}{2}{\tan }^{-1}\frac{x}{2}+\frac{9}{5}\times \frac{1}{3}{\tan }^{-1}\frac{x}{3}+c \\ =-\frac{2}{5}{\tan }^{-1}\frac{x}{2}+\frac{3}{5}{\tan }^{-1}\frac{x}{3}+c \\ \mathrm{Hence},\int \frac{x^2}{\left(x^2+4\right)\left(x^2+9\right)}\mathrm{d}x=-\frac{2}{5}{\tan }^{-1}\frac{x}{2}+\frac{3}{5}{\tan }^{-1}\frac{x}{3}+c \end{gathered}$$