Question

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

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

Answer 1

The correct option is C $\frac{1}{e}\log (x^e+e^x)+c$

Put $x^e+e^x=t$
$\Rightarrow e(x^{e-1}+e^{x-1})dx=dt$
$\therefore \int \frac{x^{e-1}+e^{x-1}}{x^e+e^x}dx=\frac{1}{e}\int \frac{dt}{t}$
$=\frac{1}{e}\log t+c$
$=\frac{1}{e}\log (x^e+e^x)+c$