Question

$\int x\text{Tan}^{-1}\sqrt{\frac{1+x^2}{1-x^2}}dx=$

• A. $\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)+\frac{1}{4}\sqrt{1-x^4}+c$
• B. $\frac{x^2}{4}(\pi-\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^4}+c$
• C. $\frac{x^2}{4}(\pi+\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^4}+c$
• D. $\frac{x^2}{4}(\pi+\text{Cos}^{-1}x^2)-\frac{1}{4}\sqrt{1-x^2}+c$

Hint:

For integrals involving terms like $\sqrt{\frac{1 \pm x^2}{1 \mp x^2}}$, the substitution $x^2 = \cos\theta$ is often very effective, as it allows for simplification using half-angle trigonometric identities.

Answer 1

The Correct Option is A