∫ 2 e ^ x + 3 e ^ - x 4 e ^ x + 7 e ^ - x dx = (1)/(14)(ux + v _ e (4 e ^ x + 7 e ^ - x )) + C , where C is a constant of integration, then u + v is equal to

Correct answer: 0 I = ∫ 2 e ^ x + 3 e ^ - x 4 e ^ x + 7 e ^ - x dx = ∫ 2 e ^ x + 3 4 e ^ 2x + 7 dx Here 2 e ^ 2x + 3 = A(8 e ^ 2x ) + B(4 e ^ 2x + 7) ⇒ 2 e ^ 2x + 3 = ( 8A + 4B ) e ^ 2x + 7B ⇒ B = (3)/(7) and A = (1)/(28) I = ∫ (1)/(28)(8 e ^ 2x ) + (3)/(7)(4 e ^ 2x + 7) 4 e ^ 2x + 7 dx I = (1)/(28) |4 e ^ 2x + 7| + (3)/(7)x + C I = (1)/(28)ln ∣ e ^ x (4 e ^ x + 7 e ^ - x ∣ + (3)/(7)x + C) ⇒ u = (13)/(2) and v = (1)/(2) ⇒ u + v = (13)/(2) + (1)/(2) = 7

JEE Main 202127 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2021 · 27 August · Evening Shift · Mathematics · Indefinite Integration

∫ 2 e ^ x + 3 e ^ - x 4 e ^ x + 7 e ^ - x dx = (1)/(14)(ux + v _ e...

∫ \dfrac{2{e}^{x} + 3{e}^{- x}}{4{e}^{x} + 7{e}^{- x}}dx = \dfrac{1}{14}\left(ux + v{\log}_{e}\left(4{e}^{x} + 7{e}^{- x}\right)\right) + C

, where

C

is a constant of integration, then

u + v

is equal to

Correct answer

Step-by-step solution

I = ∫ \dfrac{2{e}^{x} + 3{e}^{- x}}{4{e}^{x} + 7{e}^{- x}}dx = ∫ \dfrac{2{e}^{x} + 3}{4{e}^{2x} + 7}dx

Here

2{e}^{2x} + 3 = A\left(8{e}^{2x}\right) + B\left(4{e}^{2x} + 7\right)

⇒ 2{e}^{2x} + 3 = ( 8A + 4B ) {e}^{2x} + 7B

⇒ B = \dfrac{3}{7} \text{ and } A = \dfrac{1}{28}

I = ∫ \dfrac{\dfrac{1}{28}\left(8{e}^{2x}\right) + \dfrac{3}{7}\left(4{e}^{2x} + 7\right)}{4{e}^{2x} + 7}dx

I = \dfrac{1}{28}\ln \left|4{e}^{2x} + 7\right| + \dfrac{3}{7}x + C

I = \dfrac{1}{28}ln ∣ {e}^{x}\left(4{e}^{x} + 7{e}^{- x} ∣ + \dfrac{3}{7}x + C\right)<br />

⇒ u = \dfrac{13}{2} \text{ and } v = \dfrac{1}{2}

<br />

⇒ u + v = \dfrac{13}{2} + \dfrac{1}{2} = 7

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