Question

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

Answer 1

Answer: 7

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