If \(\int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}\left(u x+v \log _{e}\left(4 e^{x}+7 e^{-x}\right)\right)+C\) where \(C\) is a constant of integration, then \(u+v\) is equal to

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Indefinite Integrals

\int \frac{2 e ^{x} + 3 e ^{- x}}{4 e ^{x} + 7 e ^{- x}} d x = \frac{1}{14} (ux + v lo g_{e} (4 e^{x} + 7 e^{- x})) + C

where

C

is a constant of integration, then

u + v

is equal to

Answer: 7

Writing the numerator as a combination of the denominator and its derivative, with

D = 4 e^{x} + 7 e^{- x}

and

D^{'} = 4 e^{x} - 7 e^{- x}

, set

2 e^{x} + 3 e^{- x} = A D + B D^{'}

. Matching the

e^{x}

and

e^{- x}

terms gives

4 A + 4 B = 2

and

7 A - 7 B = 3

, so

A = \frac{13}{28}

and

B = \frac{1}{28}

. Hence

\int \frac{2 e ^{x} + 3 e ^{- x}}{4 e ^{x} + 7 e ^{- x}} d x = A x + B ln D + C = \frac{1}{14} (\frac{13}{2} x + \frac{1}{2} ln (4 e^{x} + 7 e^{- x})) + C .

Thus

u = \frac{13}{2}

v = \frac{1}{2}

, and

u + v = 7

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer