Question

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

Answer 1

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

Let $x+\frac{1}{x}=t$

$1-\frac{1}{x^2}dx=dt\Rightarrow I=\int \frac{-dt}{t^2-1}$

$=-\frac{1}{2}\int \frac{dt}{t-1}-\frac{dt}{t+1}$

$=-\frac{1}{2}ln\left(\frac{t-1}{t+1}\right)+C=\frac{1}{2}ln\sqrt{\frac{x^2+x+1}{x^2+x-1}}+C$