Question

Evaluate the following integrals:

$\int \frac{x^2}{x^4-x^2-12}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2}{x^4-x^2-12}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{x^2}{x^4-x^2-12}=\frac{x^2}{x^4-4x^2+3x^2-12} \\ =\frac{x^2}{\left(x^2-4\right)\left(x^2+3\right)} \\ =\frac{A}{x^2-4}+\frac{B}{x^2+3} \\ \Rightarrow x^2=A\left(x^2+3\right)+B\left(x^2-4\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 1=A+B\mathrm{and}0=3A-4B \\ \mathrm{or}A=\frac{4}{7}\mathrm{and}B=\frac{3}{7} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{4}{7}}}{x^2-4}+\frac{{\displaystyle \frac{3}{7}}}{x^2+3}\right)\mathrm{d}x \\ =\frac{4}{7}\int \frac{1}{x^2-4}\mathrm{d}x+\frac{3}{7}\int \frac{1}{x^2+3}\mathrm{d}x \\ =\frac{4}{7}\times \frac{1}{4}\log \left|\frac{x-2}{x+2}\right|+\frac{\sqrt{3}}{7}{\tan }^{-1}\frac{x}{\sqrt{3}}+c \\ =\frac{1}{7}\log \left|\frac{x-2}{x+2}\right|+\frac{\sqrt{3}}{7}{\tan }^{-1}\frac{x}{\sqrt{3}}+c \\ \mathrm{Hence},\int \frac{x^2}{x^4-x^2-12}\mathrm{d}x=\frac{1}{7}\log \left|\frac{x-2}{x+2}\right|+\frac{\sqrt{3}}{7}{\tan }^{-1}\frac{x}{\sqrt{3}}+c \end{gathered}$$