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