Question

Solve
$\int \frac{x^2+1}{x^4-x^2+1}dx$.

Answer 1

We have,

$\begin{array}{c}\int \frac{x^2+1}{x^4-x^2+1}dx\\ =\int \frac{1+\frac{1}{x^2}}{x^2-\frac{1}{x^2}-1}dx\\ =\int \frac{\left(1+\frac{1}{x^2}\right)}{{\left(x-\frac{1}{x}\right)}^2+1}dxx-\frac{1}{x}=t,\frac{dt}{dx}=1+\frac{1}{x^2}\\ =\int \frac{dt}{t^2+1}={\tan }^{-1}t+C\\ ={\tan }^{-1}\left(x-\frac{1}{x}\right)+C\end{array}$