Mathematics

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


ANSWER

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


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
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