Question

Evaluate : $\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e}dx.$

• A. $\frac{1}{e}\log (e^x-x^e).$
• B. $\frac{1}{e}\log (e^x+x^e).$
• C. $\log (e^x+x^e).$
• D. $\frac{1}{e^2}\log (e^x+x^e).$

Answer 1

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

Let $I=\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e}dx$

Multiply numerator and denominator by $e$
$I=\frac{1}{e}\int \frac{e{e}^{x-1}+e{x}^{e-1}}{e^x+x^e}dx=\frac{1}{e}\int \frac{e^x+e{x}^{e-1}}{e^x+x^e}dx$
Put $e^x+x^e=t\Rightarrow (e^x+e{x}^{e-1})dx=dt$

Therefore

$I=\frac{1}{e}\int \frac{dt}{t}=\frac{1}{e}\log t=\frac{1}{e}\log (e^x+x^e)$

Hence, option 'B' is correct.