Math, asked by AestheticSky, 4 hours ago

please answer it !

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

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Answers

Answered by SparklingBoy
40

▪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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Answered by BrainlyTurtle
18

Answer>>> Refer to Attachment

Explanation >>>Refer to Attachment

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