Mathematics

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

$\displaystyle \frac{1}{e}\log \left ( e^{x}+x^{e} \right ).$

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

Multiply numerator and denominator by $e$

$\displaystyle I=\frac { 1 }{ e } \int \frac { ee^{ x-1 }+ex^{ e-1 } }{ e^{ x }+x^{ e } } dx=\frac { 1 }{ e } \int \frac { e^{ x }+ex^{ e-1 } }{ e^{ x }+x^{ e } } dx$

Put $\displaystyle e^{ x }+x^{ e }=t\Rightarrow \left( e^{ x }+ex^{ e-1 } \right) dx=dt$

Therefore

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

Hence, option 'B' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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