Evalute the following integrals: e^ x-1 +x^ e-1 e^x+x^e dx - Vidyadip

Question

Evalute the following integrals:
$\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
Putting e^x + x^e = t
$\Rightarrow\text{e}^\text{x}+\text{ex}^{\text{e}-1}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)\text{dx}=\frac{\text{dt}}{\text{e}}$
$\therefore\text{I}=\frac{1}{\text{e}}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{\text{e}}\text{ ln}|\text{t}|+\text{C}$
$=\frac{1}{\text{e}}\text{ ln}\Big|\text{e}^\text{x}+\text{x}^\text{e}\Big|+\text{C}\big[\because\text{t}=\text{e}^\text{x}+\text{x}^\text{e}\big]$

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