Question

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

Answer 1

$\int \frac{2 e ^{ x }}{4 e ^{ x }+7 e ^{- x }} dx +3 \int \frac{ e ^{- x }}{4 e ^{ x }+7 e ^{- x }} dx$
$=\int \frac{2 e ^{2 x }}{4 e ^{2 x }+7} dx +3 \int \frac{ e ^{-2 x }}{4+7 e ^{-2 x }} dx$
Let $4 e ^{2 x }+7= T$
$8 e ^{2 x } dx = dT$
$2 e ^{2 x } dx =\frac{ dT }{4}$
Let $4+7 e ^{-2 x }= t $
$-14 e ^{-2 x } dx = dt $
$e ^{-2 x } dx =-\frac{ dt }{14}$
$\int \frac{ dT }{4 T }-\frac{3}{14} \int \frac{ dt }{ t }$
$=\frac{1}{4} \log T -\frac{3}{14} \log t + C$
$=\frac{1}{4} \log \left(4 e ^{2 x }+7\right)-\frac{3}{14} \log \left(4+7 e ^{-2 x }\right)+ C$
$=\frac{1}{14}\left[\frac{1}{2} \log \left(4 e ^{ x }+7 e ^{- x }\right)+\frac{13}{2} x \right]+ C$
$u =\frac{13}{2}, v =\frac{1}{2} $
$\Rightarrow u + v =7$
Aliter :
$2 e ^{ x }+3 e ^{- x }= A \left(4 e ^{ x }+7 e ^{- x }\right)+ B \left(4 e ^{ x }-7 e ^{- x }\right)+\lambda$
$2=4 A +4 B ; 3=7 A -7 B ; \lambda=0 $
$ A + B =\frac{1}{2} $
$A - B =\frac{3}{7}$
$ A =\frac{1}{2}\left(\frac{1}{2}+\frac{3}{7}\right)=\frac{7+6}{28}=\frac{13}{28}$
$ B = A -\frac{3}{7}=\frac{13}{28}-\frac{3}{7}=\frac{13-12}{28}=\frac{1}{28}$
$\int \frac{13}{28} dx +\frac{1}{28} \int \frac{4 e ^{ x }-7 e ^{- x }}{4 e ^{ x }+7 e ^{- x }} dx$
$\frac{13}{28} x +\frac{1}{28} \ell n \left|4 e ^{ x }+7 e ^{- x }\right|+ C$
$u =\frac{13}{2} ; v =\frac{1}{2}$
$\Rightarrow u + v =7$