JEE Main 20203 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2020 · 3 September · Evening Shift · Mathematics · Indefinite Integration

If ∫ ^ - 1 ( x 1 + x )dx = A(x) ^ - 1 ( x ) + B(x) + C , where C is...

∫ \sin^{- 1}\left(\dfrac{\sqrt{x}}{1 + x}\right)dx = A\left(x\right)\tan^{- 1}\left(\sqrt{x}\right) + B\left(x\right) + C

, where

C

is a constant of integration, then the ordered pair \left(A\left(x\right), B\left(x\right)\right)$ can be :

Options

A $\left(x - 1, \sqrt{x}\right)$
B $\left(x - 1, - \sqrt{x}\right)$
C $\left(x + 1, \sqrt{x}\right)$
D $\left(x + 1, - \sqrt{x}\right)$

Correct answer

D. $\left(x + 1, - \sqrt{x}\right)$

Step-by-step solution

I = ∫ \sin^{- 1}\left(\dfrac{\sqrt{x}}{1 + x}\right)dx $$ = ∫ \mathop{\mathop{\tan^{- 1}\left(\sqrt{x}\right)}_{⏟}}_{I}\mathop{\mathop{\left(1\right)}_{⏟}}_{II}dx $$ = x\tan^{- 1}\sqrt{x} - ∫ \dfrac{1}{1 + x} . \dfrac{1}{2\sqrt{x}} . xdx + C $$ = x\tan^{- 1}\sqrt{x} - \dfrac{1}{2} ∫ \dfrac{t . 2t . dt}{1 + {t}^{2}} + C ( $putting$ x = {t}^{2} ⇒ dx = 2tdt ) $$ = x\tan^{- 1}\sqrt{x} - ∫ \dfrac{{t}^{2}}{1 + {t}^{2}}dt + C $$ = x\tan^{- 1}\sqrt{x} - t + \tan^{- 1}t + C $$ = x\tan^{- 1}\sqrt{x} - \sqrt{x} + \tan^{- 1}\sqrt{x} + C $$ = \left(x + 1\right)\tan^{- 1}\sqrt{x} - \sqrt{x} + C $Comparing with$ A\left(x\right)\tan^{- 1}\left(\sqrt{x}\right) + B\left(x\right) + C $, we get\left(A\left(x\right), B\left(x\right)\right) = \left(x + 1, - \sqrt{x}\right)$

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