JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Indefinite Integrals

\int sin^{- 1} (\frac{x}{1 + x}) d x = A (x) tan^{- 1} (x) + B (x) + C

, where

C

is a constant of integration, then the ordered pair

(A (x), B (x))

can be:

Answer: (C)

Solution

Since

sin^{- 1} \frac{x}{1 + x} = tan^{- 1} x

, integrate

\int tan^{- 1} x d x

by parts:

x tan^{- 1} x - \int x \cdot \frac{1}{1 + x} \cdot \frac{1}{2 x} d x = x tan^{- 1} x - \int \frac{x}{2 ( 1 + x )} d x

. Putting

t = x

, the remaining integral is

\int \frac{t ^{2}}{1 + t ^{2}} d t = t - tan^{- 1} t = x - tan^{- 1} x

. So the result is

x tan^{- 1} x - (x - tan^{- 1} x) + C = (x + 1) tan^{- 1} x - x + C

, giving

A (x) = x + 1

and

B (x) = - x

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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