Question

please answer it !

$\large \displaystyle \sf \int \frac{ {e}^{x - 1} + {x}^{e - 1} }{ {e}^{x} + {x}^{e} }$

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Answer 1

▪To Find :-

$\large \displaystyle\sf \int \frac{ {e}^{x - 1} + {x}^{e - 1} }{ {e}^{x} + {x}^{e} }dx$

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▪Formulae Used :-

$\maltese \: \: \: \bf \dfrac{d}{dx} \: {e}^{x} = {e}^{x} \\ \\ \maltese \: \: \: \bf \frac{d}{dx} \: x {}^{n} = nx {}^{n - 1} \\ \\ \maltese \: \: \: \bf \int \frac{1}{x} \: dx = ln|x| + C$

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▪Solution :-

$\bf Put: - ) \: {e}^{x} + {x}^{e} = t \\ \\ \implies \sf \{ {e}^{x} + ex {}^{(e - 1)} \}dx = dt \\ \\ \bf \: \large\{ Taking \: \: e \: \: common \} \\ \\ \sf e( {e}^{x - 1} + {x}^{e - 1} )dx = dt \\ \\ \implies \bf ( {e}^{x - 1} + {x}^{e - 1} )dx = \frac{1}{e} \: dt$

Now ,

$\large \sf \int \frac{ {e}^{x - 1} + {x}^{e - 1} }{ {e}^{x} + {x}^{e} }dx \\ \\ = \sf \int \frac{1}{t} . \frac{1}{e} dt \\ \\ = \sf \frac{1}{e} \int \frac{dt}{t} \\ \\ = \sf \frac{1}{e} (ln \:| t|) + C$

{Putting Back The Value of t}

$\large \purple{ = \underline {\boxed{ \bf \frac{1}{e} \bigg \{ln|{e}^{x} + {x}^{e} |\bigg \} + C}} }$

Where C is Constant of Integration.

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Answer>>> Refer to Attachment

Explanation >>>Refer to Attachment