$$ \int x \tan ^{-1} \sqrt{\frac{1+x^2}{1-x^2}} d x= $$
- A
$\frac{x^2}{4}\left(\pi-\cos ^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^2}+C$
- B
$\frac{x^2}{4}\left(\pi-\cos ^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^4}+C$
- C
$\frac{x^2}{4}\left(\pi+\cos ^{-1} x^2\right)-\frac{1}{4} \sqrt{1-x^4}+C$
- D
$\frac{x^2}{4}\left(\pi+\cos ^{-1} x^2\right)-\frac{1}{4} \sqrt{1-x^2}+C$
✓ Correct answer: B
Want the full step-by-step reasoning?