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

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(C) Let $I = \int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} dx$ ... $(i)$ Put $x^e + e^x = t$ ... $(ii)$ Differentiating both sides with respect to $x$,we get:

Vedclass · Accepted Answer

$\frac{1}{e} \log \left|x^e+e^x\right|+C$

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(C) Let $I = \int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} dx$ ... $(i)$ Put $x^e + e^x = t$ ... $(ii)$ Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(x^e + e^x) = \frac{dt}{dx}$ $e x^{e-1} + e^x = \frac{dt}{dx}$ $(e x^{e-1} + e^x) dx = dt$ Factor out $e$: $e(x^{e-1} + \frac{e^x}{e}) dx = dt$ $e(x^{e-1} + e^{x-1}) dx = dt$ $(x^{e-1} + e^{x-1}) dx = \frac{1}{e} dt$ ... $(iii)$ Substituting $(ii)$ and $(iii)$ into $(i)$: $I = \int \frac{1}{t} \cdot \frac{1}{e} dt$ $I = \frac{1}{e} \int \frac{1}{t} dt$ $I = \frac{1}{e} \log |t| + C$ Substituting back $t = x^e + e^x$: $I = \frac{1}{e} \log |x^e + e^x| + C$

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$

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