$\int \frac{x^2-1}{x \sqrt{x^4+3 x^2+1}} d x$ is equal to
Answer & explanation
Correct answer: option 1
We have,
$I =\int \frac{x^2-1}{x \sqrt{x^4+3 x^2+1}} d x=\int \frac{1-\frac{1}{x^2}}{\sqrt{x^2+3+\frac{1}{x^2}}} d x$
$\Rightarrow I =\int \frac{1}{\sqrt{\left(x+\frac{1}{x}\right)^2+1^2}} d\left(x+\frac{1}{x}\right)$
$\Rightarrow I =\log _e \mid x+\frac{1}{x}+\sqrt{\left(x+\frac{1}{x}\right)^2+1 \mid+C}$