int frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x is | vedclass

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(C) Let $I = \int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x$. Substitute $t = x^{e}+e^{x}$. Differentiating both sides with respect to $x$,we get $dt =

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x$. Substitute $t = x^{e}+e^{x}$. Differentiating both sides with respect to $x$,we get $dt = (e x^{e-1} + e^{x-1} \cdot \ln(e)) dx$. Since $\ln(e) = 1$,this simplifies to $dt = e(x^{e-1} + e^{x-1}) dx$. Therefore,$(x^{e-1} + e^{x-1}) dx = \frac{1}{e} dt$. Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{1}{e} dt = \frac{1}{e} \int \frac{1}{t} dt$. Integrating,we get $I = \frac{1}{e} \log|t| + c$. Substituting back $t = x^{e}+e^{x}$,we get $I = \frac{1}{e} \log(x^{e}+e^{x}) + c$.

$\int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x$ is equal to

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