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Integral with arcsin + arctan for x > 1 | JEE

JEE Maths question with a full step-by-step solution.

Question
For x>1x>1, ∫11−x8{sin⁡−1 ⁣(2x1+x2)+tan⁡−1 ⁣(2x1−x2)}dx\displaystyle\int \dfrac{1}{1-x^{8}}\left\{\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)+\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right)\right\}dx is
Aconstantcorrect
Bπ8tan⁡−1x+c\dfrac{\pi}{8}\tan^{-1}x+c
C−π8sin⁡−1x+c-\dfrac{\pi}{8}\sin^{-1}x+c
Dπ2+x+c\dfrac{\pi}{2}+x+c
Solution
Step 1: Evaluate each inverse-trig term for x>1x>1. The double-angle forms of sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1} leave their principal ranges once x>1x>1, giving the reduced values
sin⁡−1 ⁣(2x1+x2)=π−2tan⁡−1x,tan⁡−1 ⁣(2x1−x2)=−π+2tan⁡−1x\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)=\pi-2\tan^{-1}x,\qquad \tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right)=-\pi+2\tan^{-1}x
 Step 2: Add the two expressions inside the braces:
(π−2tan⁡−1x)+(−π+2tan⁡−1x)=0\big(\pi-2\tan^{-1}x\big)+\big(-\pi+2\tan^{-1}x\big)=0
 Step 3: The bracketed factor is identically zero, hence
I=∫11−x8⋅0 dx=∫0 dx=cI=\int\dfrac{1}{1-x^{8}}\cdot 0\,dx=\int 0\,dx=c
 The integral is a constant. Correct answer: (1)
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