If ∫ ( e ^ 2x + 2 e ^ x - e ^ - x - 1) e ^ ( e ^ x + e ^ - x ) dx = g ( x ) e ^ ( e ^ x + e ^ - x ) + c, where c is a constant of integration, then g ( 0 ) is

Correct answer: D. 2 I = ∫ ( e ^ 2x + 2 e ^ x - e ^ - x - 1) e ^ ( e ^ x + e ^ - x ) dx ⇒ I = ∫ ( e ^ 2x + e ^ x - 1) e ^ ( e ^ x + e ^ - x ) dx + ∫ ( e ^ x - e ^ - x ) e ^ e ^ x + e ^ - x dx ⇒ I = ∫ ( e ^ x + 1 - e ^ - x ) e ^ e ^ x + e ^ - x + x dx + e ^ e ^ x + e ^ - x + c e ^ x + e ^ - x + x = u ( e ^ x - e ^ - x + 1)dx = du I = e ^ e ^ x + e ^ - x + x + e ^ e ^ x + e ^ - x + c = e ^ e ^ x + e ^ - x ( e ^ x + 1) + c then g ( x ) = e ^ x + 1 g ( 0 ) = 2

JEE Main 20205 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 5 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2020 · 5 September · Morning Shift · Mathematics · Indefinite Integration

If ∫ ( e ^ 2x + 2 e ^ x - e ^ - x - 1) e ^ ( e ^ x + e ^ - x ) dx =...

∫ \left({e}^{2x} + 2{e}^{x} - {e}^{- x} - 1\right){e}^{\left({e}^{x} + {e}^{- x}\right)}dx = g ( x ) {e}^{\left({e}^{x} + {e}^{- x}\right)} + c,

where

c

is a constant of integration, then

g ( 0 )

Options

A $e$
B ${e}^{2}$
C $1$
D $2$

Correct answer

D. $2$

Step-by-step solution

I = ∫ \left({e}^{2x} + 2{e}^{x} - {e}^{- x} - 1\right){e}^{\left({e}^{x} + {e}^{- x}\right)}dx $$ ⇒ I = ∫ \left({e}^{2x} + {e}^{x} - 1\right){e}^{\left({e}^{x} + {e}^{- x}\right)}dx + ∫ \left({e}^{x} - {e}^{- x}\right){e}^{{e}^{x} + {e}^{- x}}dx $$ ⇒ I = ∫ \left({e}^{x} + 1 - {e}^{- x}\right){e}^{{e}^{x} + {e}^{- x} + x}dx + {e}^{{e}^{x} + {e}^{- x}} + c $$ {e}^{x} + {e}^{- x} + x = u $\left({e}^{x} - {e}^{- x} + 1\right)dx = du$

$I = {e}^{{e}^{x} + {e}^{- x} + x} + {e}^{{e}^{x} + {e}^{- x}} + c = {e}^{{e}^{x} + {e}^{- x}}\left({e}^{x} + 1\right) + c$

then $g ( x ) = {e}^{x} + 1$

$g ( 0 ) = 2$

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