Question

Evaluate the following integrals:

$\int \frac{2x^2+1}{x^2\left(x^2+4\right)}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{2x^2+1}{x^2\left(x^2+4\right)}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{2x^2+1}{x^2\left(x^2+4\right)}=\frac{A}{x^2}+\frac{B}{x^2+4} \\ \Rightarrow 2x^2+1=A\left(x^2+4\right)+B\left(x^2\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 2=A+B\mathrm{and}1=4A \\ \mathrm{or}A=\frac{1}{4}\mathrm{and}B=\frac{7}{4} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{1}{4}}}{x^2}+\frac{{\displaystyle \frac{7}{4}}}{x^2+4}\right)\mathrm{d}x \\ =\frac{1}{4}\int \frac{1}{x^2}\mathrm{d}x+\frac{7}{4}\int \frac{1}{x^2+4}\mathrm{d}x \\ =-\frac{1}{4x}+\frac{7}{8}{\tan }^{-1}\frac{x}{2}+c \\ \mathrm{Hence},\int \frac{2x^2+1}{x^2\left(x^2+4\right)}\mathrm{d}x=-\frac{1}{4x}+\frac{7}{8}{\tan }^{-1}\frac{x}{2}+c \end{gathered}$$