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)

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