Question

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

Answer 1

$=\frac{1}{2}\int \frac{(x^2+1)-(x^2-1)dx}{x^4+x^2+1}$

$=\frac{1}{2}\int \frac{(x^2+1)dx}{x^4+x^2+1}-\int \frac{(x^2-1)dx}{x^4+x^2+1})$

$=\frac{1}{2}\int \frac{(1+\frac{1}{x^2})}{x^2+1+\frac{1}{x^2}}-\int \frac{(1-\frac{1}{x^2})dx}{x^2+1+\frac{1}{x^2}}$

$=\frac{1}{2}\int \frac{(1+\frac{1}{x^2})}{(x-\frac{1}{x}{)}^2+(\sqrt{3}{)}^2}-\int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x}{)}^2-1^2}$

$=\frac{1}{2}\left[\frac{1}{\sqrt{3}}{\tan }^{-1}\right(\frac{x-\frac{1}{x}}{\sqrt{3}})-\frac{1}{2}\log \left(\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right)+c]$