$\smallint \,\,\frac{{{x^{e - 1}}\,\, + \,\,{e^{x - 1}}}}{{{x^e}\,\, + \,\,{e^x}}}\;\,\,dx$ is equal to

$\log \,\,\left( {{x^e}\,\, + \,\,{e^x}} \right) + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Understanding +4 | -1

\int \,\,\frac{{{x^{e - 1}}\,\, + \,\,{e^{x - 1}}}}{{{x^e}\,\, + \,\,{e^x}}}\;\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{1}{e}\,\,\log \,\,({x^e}\,\, + \,\,{e^x}) + c

Option C,

\frac{1}{e}\,\log \,({x^e}\,+\,{e^x}) + c

, is correct because it results from a u-substitution where the derivative of the denominator is proportional to the numerator.