Question

If $\int \frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$ is equal to
$\frac{\sqrt{2}}{4}A{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\sqrt{x^2+1/x^2}\right)$+C then A is equal to.

Answer 1

$I=\int \frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}dx$
Substituting $t=x+\frac{1}{x}\Rightarrow dt=(1-\frac{1}{x^2})dx$, we get
$I=\int \frac{dt}{t\sqrt{t^2-2}}=\int \frac{tdt}{t^2\sqrt{t^2-2}}$
Again substituting $y^2=t^2-2$, we get
$I=\int \frac{ydy}{(y^2+2)y}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{y}{\sqrt{2}}+C=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}\left(\sqrt{{(x+\frac{1}{x})}^2-2}\right)+C=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\sqrt{x^2+\frac{1}{x^2}}\right)+C\Rightarrow A=2$