Question

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x}dx$ is equal to

• A. $\frac{1}{e}log\left(x^e-e^x\right)+c$
• B. $\frac{1}{e}log\left(x^e+e^x\right)+c$
• C. $\frac{1}{e}log\left(x^x-e^e\right)+c$
• D. None of the above

Answer 1

Answer: (B)

$I=\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x}dx$
$Let{x}^e+e^x=t$
$\Rightarrow \left(e{x}^{e-1}+e^x\right)dx=dt$
$\Rightarrow e\left(x^{e-1}+\frac{e^x}{e}\right)dx=dt$
$\Rightarrow e\left(x^{e-1}+e^{x-1}\right)dx=dt$
$\Rightarrow \left(x^{e-1}+e^{x-1}\right)dx=\frac{dt}{e}$
$\therefore I=\int \frac{dt}{e\cdot t}=\frac{1}{e}\int \frac{dt}{t}=\frac{1}{e}\log t+c$
$=\frac{1}{e}\log \left(x^e+e^x\right)+c$