Question

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

Answer 1

Let $I=\int \frac{e^{\log (1+1/x^2)}}{x^2+1/x^2}dx=\int \frac{(1+1/x^2)}{2+{(x-\frac{1}{x})}^2}dx$
Put $x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^2})dx=dt$
$\therefore I=\int \frac{dt}{2+t^2}=\frac{1}{\sqrt{2}}{\tan }^{-1}t=\frac{1}{\sqrt{\left(2\right)}}{\tan }^{-1}(x-\frac{1}{x})$