Question

Evaluate the following integrals:

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

Answer 1

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