Question
$\text { If } \int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}\left(u x+v \log _{c}\left(4 e^{x}+7 e^{-x}\right)\right)+C$ where $\mathrm{C}$ is a constant of integration, then $\mathrm{u}+\mathrm{v}$ is equal to .... .
$=\int \frac{2 e^{2 x}}{4 e^{2 x}+7} d x+3 \int \frac{e^{-2 x}}{4+7 e^{-2 x}} d x$
Let $\quad 4 e^{2 x}+7=T \quad$ Let $\quad 4+7 e^{-2 x}=t$
$8 e^{2 x} d x=d T \quad\quad\quad\quad-14 e^{-2 x} d x=d t$
$2 e^{2 x} d x=\frac{d T}{4} \quad\quad\quad\quad e^{-2 x} d x=-\frac{d t}{14}$
$\int \frac{d T}{4 T}-\frac{3}{14} \int \frac{d t}{t}$
$=\frac{1}{4} \log T-\frac{3}{14} \log t+C$
$=\frac{1}{4} \log \left(4 e^{2 x}+7\right)-\frac{3}{14} \log \left(4+7 e^{-2 x}\right)+C$
$=\frac{1}{14}\left[\frac{1}{2} \log \left(4 e^{x}+7 e^{-x}\right)+\frac{13}{2} x\right]+C$
$u=\frac{13}{2}, v=\frac{1}{2} \Rightarrow u+v=7$
Aliter':
$2 \mathrm{e}^{\mathrm{x}}+3 \mathrm{e}^{-x}=\mathrm{A}\left(4 \mathrm{e}^{\mathrm{x}}+7 \mathrm{e}^{-x}\right)+\mathrm{B}\left(4 \mathrm{e}^{\mathrm{x}}-7 \mathrm{e}^{-x}\right)+\lambda$
$2=4 \mathrm{~A}+4 \mathrm{~B} ; 3=7 \mathrm{~A}-7 \mathrm{~B} \quad ; \quad \lambda=0$
$\mathrm{~A}+\mathrm{B}=\frac{1}{2}$
$\mathrm{~A}-\mathrm{B}=\frac{3}{7}$
$\mathrm{~A}=\frac{1}{2}\left(\frac{1}{2}+\frac{3}{7}\right)=\frac{7+6}{28}=\frac{13}{28}$
$\mathrm{~B}=\mathrm{A}-\frac{3}{7}=\frac{13}{28}-\frac{3}{7}=\frac{13-12}{28}=\frac{1}{28}$
$\int \frac{13}{28}\, \mathrm{~dx}+\frac{1}{28} \int \frac{4 \mathrm{e}^{\mathrm{x}}-7 \mathrm{e}^{-\mathrm{x}}}{4 \mathrm{e}^{\mathrm{x}}+7 \mathrm{e}^{-\mathrm{x}}}\, \mathrm{dx}$
$\frac{13}{28} \mathrm{x}+\frac{1}{28} \ln \left|4 \mathrm{e}^{\mathrm{x}}+7 \mathrm{e}^{-\mathrm{x}}\right|+\mathrm{C}$
$\mathrm{u}=\frac{13}{2} ; \mathrm{v}=\frac{1}{2}$
$\Rightarrow \mathrm{u}+\mathrm{v}=7$
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