Question

$\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ is equal to

• A. $\sqrt{\frac{x^2+1}{x^2-1}}+c$
• B. $\sqrt{\frac{x^2-1}{x^2+1}}+c$
• C. $\sqrt{x^4-1}+c$
• D. None of these

Answer 1

The correct option is A

$\sqrt{\frac{x^2+1}{x^2-1}}+c$

$\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}$
$=\int \frac{-2x}{(x^2-1{)}^{3/2}\sqrt{x^2+1}}$
Put $\sqrt{\frac{x^2+1}{x^2-1}}=z$
$\therefore dz=dx\frac{1}{2}{\left(\frac{x^2-1}{x^2+1}\right)}^{1/2}.\frac{(x^2-1).2x-(x^2+1)2x}{(x^2-1{)}^2}$

$\Rightarrow \frac{\sqrt{x^2-1}}{\sqrt{x^2+1}}.\frac{-2x}{(x^2-1{)}^2}dx=dz$
$\Rightarrow \frac{-2x}{(x^2-1{)}^{3/2}\sqrt{x^2+1}}dx=dz$
$\therefore \text{Given integral}=\int dz=z+c=\sqrt{\frac{x^2+1}{x^2-1}}+c$