Question

please give me solution of this​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\dfrac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}}\,dx$

$\displaystyle=\int\dfrac{x^{e-1}+\dfrac{e^{x}}{e}}{x^{e}+e^{x}}\,dx$

$\displaystyle=\int\dfrac{\dfrac{e\,x^{x-1}+e^{x}}{e}}{x^{e}+e^{x}}\,dx$

$\displaystyle=\dfrac{1}{e}\int\dfrac{e\,x^{x-1}+e^{x}}{x^{e}+e^{x}}\,dx$

$\bf{Put\,\,\,\,x^{e}+e^{x}=t}$

$\bf{\implies\,\left(e\,x^{e-1}+e^{x}\right)dx=dt}$

$\displaystyle=\dfrac{1}{e}\int\dfrac{dt}{t}$

$\displaystyle=\dfrac{1}{e}\,\ln|t|+C$

$\displaystyle=\dfrac{1}{e}\,\ln|x^e+e^x|+C$

Answer 2

Given Question

Evaluate the following

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{e - 1} + {e}^{x - 1} }{ {x}^{e} + {e}^{x} } \: dx$

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{e - 1} + {e}^{x - 1} }{ {x}^{e} + {e}^{x} } \: dx$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm :\longmapsto\: {x}^{e} + {e}^{x} = y$

$\rm :\longmapsto\: \dfrac{d}{dx}({x}^{e} + {e}^{x}) = \dfrac{d}{dx}y$

$\rm :\longmapsto\: {ex}^{e - 1} + {e}^{x} = \dfrac{dy}{dx}$

$\rm :\longmapsto\:e( {x}^{e - 1} + {e}^{x - 1}) = \dfrac{dy}{dx}$

$\rm :\longmapsto\:({x}^{e - 1} + {e}^{x - 1})dx = \dfrac{dy}{e}$

So, on substituting all these values in above integral, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{dy}{e \: y}$

$\rm \:  =  \:\dfrac{1}{e} \displaystyle\int\rm \frac{dy}{ y}$

$\rm \:  =  \:\dfrac{1}{e}log |y| + c$

$\rm \:  =  \:\dfrac{1}{e}log \bigg| {x}^{e} + {e}^{x} \bigg| + c$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm \frac{ {x}^{e - 1} + {e}^{x - 1} }{ {x}^{e} + {e}^{x} } \: dx = \frac{1}{e}log \bigg| {x}^{e} + {e}^{x}\bigg| + c}}$

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LEARN MORE :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$