Question

Prove that $\int \frac{(x^2+1)dx}{(x^2+2)(2x^2+1)}=\frac{1}{3\sqrt{\left(2\right)}}\{{\tan }^{-1}\frac{x}{\sqrt{\left(2\right)}}+{\tan }^{-1}x\sqrt{2}\}.$

Answer 1

Let $I=\int \frac{(x^2+1)dx}{(x^2+2)(2x^2+1)}=\int (\frac{1}{3(2x^2+1)}+\frac{1}{3(x^2+2)})dx$
$\Rightarrow I=I_1+I_2$
Where
$I_1=\frac{1}{3}\int \frac{1}{(2x^2+1)}dx$
Put $t=\sqrt{2}x\Rightarrow dt=\sqrt{2}dx$
Therefore
$I_1=\frac{1}{3\sqrt{2}}\int \frac{1}{t^2+1}dt=\frac{1}{3\sqrt{2}}{\tan }^{-1}t=\frac{1}{3\sqrt{2}}{\tan }^{-1}\sqrt{2}x$
And
$I_2=\frac{1}{3}\int \frac{1}{x^2+2}dx=\frac{1}{3}\int \frac{1}{2(\frac{x^2}{2}+1)}dx=\frac{1}{6}\int \frac{1}{\frac{x^2}{2}+1}dx$
Put $u=\frac{x}{\sqrt{2}}\Rightarrow du=\frac{1}{\sqrt{2}}dx$
Therefore
$I_2=\frac{1}{3\sqrt{2}}\int \frac{1}{u^2+1}du=\frac{1}{3\sqrt{2}}{\tan }^{-1}u=\frac{1}{3\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}$
Hence
$I=\frac{1}{3\sqrt{\left(2\right)}}\{{\tan }^{-1}\frac{x}{\sqrt{\left(2\right)}}+{\tan }^{-1}x\sqrt{2}\}$