Question

$\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx$ is equal to

• A. $\frac{3}{2}x+\frac{35}{36}log\left|9e^{2x}-4\right|+C$
• B. $\frac{3}{2}x-\frac{35}{36}log\left|9e^{2x}-4\right|+C$
• C. $-\frac{3}{2}x+\frac{35}{36}log\left|9e^{2x}-4\right|+C$
• D. $-\frac{5}{2}x+\frac{35}{36}log\left|9e^{2x}-4\right|+C$
• E. $\frac{5}{2}x+\frac{35}{36}log\left|9e^{2x}-4\right|+C$

Answer 1

Answer: (C)

$\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx$
$=\int \frac{4e^{2x}+6}{9e^{2x}-4}dx$
$=4\int \frac{e^{2x}}{9e^{2x}-4}dx+6\int \frac{e^{-2x}}{9-4e^{-2x}}dx$
Put $t_1=9e^{2x}-4$ and $t_2=9-4e^{-2x}$
$d{t}_1=18e^{2x}dx$ and $d{t}_2=8e^{-2x}dx$
$=4\int \frac{1}{t_1}\cdot \frac{d{t}_1}{18}+6\int \frac{1}{t_2}\cdot \frac{d{t}_2}{8}$
$=\frac{2}{9}\int \frac{d{t}_1}{t_1}+\frac{3}{4}\int \frac{d{t}_2}{t_2}$
$=\frac{2}{9}\log t_1+\frac{3}{4}\log t_2+C$
$=\frac{2}{9}\log \left|9e^{2x}-4\right|+\frac{3}{4}\log \left|9-4e^{-2x}\right|+C$
$=\frac{2}{9}\log \left|9e^{2x}-4\right|+\frac{3}{4}\log \left|9e^{2x}-4\right|$
$-\frac{3}{4}\log \left|e^{2x}\right|+C$
$=\left(\frac{2}{9}+\frac{3}{4}\right)\log \left|9e^{2x}-4\right|-\frac{3}{4}(2x)+C$
$(\because \log e=1)$
$=-\frac{3}{2}x+\frac{35}{36}\log \left|9e^{2x}-4\right|+C$