Question

$\int \frac{4e^x}{2e^x-5e^{-x}}dx=$dX is equal to

• A. $4log\left|e^x-5\right|+C$
• B. $\frac{1}{4}log\left|e^{2x}-5\right|+C$
• C. $log\left|2e^x-5e^{-x}\right|+C$
• D. $4log\left|2e^x-5\right|+C$
• E. $log\left|2e^{2x}-5\right|+C$

Answer 1

Answer: (E)

Let $I=\int \frac{4e^x}{2e^x-5e^{-x}}dx$
$=\int \frac{4e^x}{2e^x-\frac{5}{e^x}}dx$
$=\int \frac{4e^{2x}}{2e^{2x}-5}dx$
$\mathrm{Put}2e^{2x}-5=t$
$\Rightarrow 4e^{2x}dx=dt$
$\therefore I=\int \frac{dt}{t}$
$=\log |t|+C$
$\Rightarrow I=\log \left|2e^{2x}-5\right|+C$