Question

Evaluate the following integrals:

$\int \frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+25} \\ \Rightarrow x^2+1=\left(Ax+B\right)\left(x^2+25\right)+\left(Cx+D\right)\left(x^2+4\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^3,x^2,x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 0=A+C\mathrm{and}1=B+D\mathrm{and}0=25A+4C\mathrm{and}1=25B+4D \\ \mathrm{or}A=0\mathrm{and}B=-\frac{1}{7}\mathrm{and}C=0\mathrm{and}D=\frac{8}{7} \\ \therefore I=\int \left(\frac{{\displaystyle -\frac{1}{7}}}{x^2+4}+\frac{{\displaystyle \frac{8}{7}}}{x^2+25}\right)\mathrm{d}x \\ =-\frac{1}{7}\int \frac{1}{x^2+4}\mathrm{d}x+\frac{8}{7}\int \frac{1}{x^2+25}\mathrm{d}x \\ =-\frac{1}{7}\times \frac{1}{2}{\tan }^{-1}\frac{x}{2}+\frac{8}{7}\times \frac{1}{5}{\tan }^{-1}\frac{x}{5}+c \\ =-\frac{1}{14}{\tan }^{-1}\frac{x}{2}+\frac{8}{35}{\tan }^{-1}\frac{x}{5}+c \\ \mathrm{Hence},\int \frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}\mathrm{d}x=-\frac{1}{14}{\tan }^{-1}\frac{x}{2}+\frac{8}{35}{\tan }^{-1}\frac{x}{5}+c \end{gathered}$$A