The integral ∫ e ^ 3 _ e 2x + 5 e ^ 2 _ e 2x e ^ 4 _ e x + 5 e ^ 3 _ e x - 7 e ^ 2 _ e x dx, x > 0 , is equal to (where c is a constant of integration)

Correct answer: B. 4 _ e | x ^ 2 + 5x - 7| + c ∫ e ^ 3 _ e 2x + 5 e ^ 2 _ e 2x e ^ 4 _ e x + 5 e ^ 3 _ e x - 7 e ^ 2 _ e x dx, x > 0 = ∫ (2x) ^ 3 + 5 (2x) ^ 2 x ^ 4 + 5 x ^ 3 - 7 x ^ 2 dx = ∫ 4 x ^ 2 (2x + 5) x ^ 2 ( x ^ 2 + 5x - 7) dx = 4 ∫ d( x ^ 2 + 5x - 7) ( x ^ 2 + 5x - 7) = 4 _ e | x ^ 2 + 5x - 7| + c

JEE Main 202125 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Indefinite Integration

The integral ∫ e ^ 3 _ e 2x + 5 e ^ 2 _ e 2x e ^ 4 _ e x + 5 e ^ 3...

The integral

∫ \dfrac{{e}^{3\log_{e}2x} + 5{e}^{2\log_{e}2x}}{{e}^{4\log_{e}x} + 5{e}^{3\log_{e}x} - 7{e}^{2\log_{e}x}}dx, x > 0

, is equal to
(where

c

is a constant of integration)

Options

A $\log_{e}\left|{x}^{2} + 5x - 7\right| + c$
B $4\log_{e}\left|{x}^{2} + 5x - 7\right| + c$
C $\dfrac{1}{4}\log_{e}\left|{x}^{2} + 5x - 7\right| + c$
D $\log_{e}\sqrt{{x}^{2} + 5x - 7} + c$

Correct answer

B. $4\log_{e}\left|{x}^{2} + 5x - 7\right| + c$

Step-by-step solution

$∫ \dfrac{{e}^{3\log_{e}2x} + 5{e}^{2\log_{e}2x}}{{e}^{4\log_{e}x} + 5{e}^{3\log_{e}x} - 7{e}^{2\log_{e}x}}dx, x > 0$

$= ∫ \dfrac{{\left(2x\right)}^{3} + 5{\left(2x\right)}^{2}}{{x}^{4} + 5{x}^{3} - 7{x}^{2}}dx = ∫ \dfrac{4{x}^{2}\left(2x + 5\right)}{{x}^{2}\left({x}^{2} + 5x - 7\right)}dx$

$= 4 ∫ \dfrac{d\left({x}^{2} + 5x - 7\right)}{\left({x}^{2} + 5x - 7\right)} = 4\log_{e}\left|{x}^{2} + 5x - 7\right| + c$

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