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What is $\int \frac{\left(x^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}-1}\right) \mathrm{dx}}{\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}}$ equal to?
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Verified Answer
The correct answer is:
$\frac{1}{\mathrm{e}} \operatorname{In}\left(\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$
$\int \frac{\left(x^{e-1}+e^{x-1}\right) d x}{x^{e}+e^{x}}$
Put $x^{e}+e^{x}=t \Rightarrow e x^{e-1}+e^{x}=\frac{d t}{d x}$
$\Rightarrow\left(e x^{e-1}+e^{x}\right) d x=d t$
$\therefore \int \frac{\left(x^{e-1}+e^{x-1}\right) d x}{x^{e}+e^{x}}=\frac{1}{e} \int \frac{e x^{e-1}+e^{x}}{x^{e}+e^{x}} \cdot d x$
$=\frac{1}{e} \int \frac{d t}{t}=\frac{1}{e} \cdot \ln t+c$
$=\frac{1}{e} \cdot \ln \left(x^{e}+e^{x}\right)+c$
Put $x^{e}+e^{x}=t \Rightarrow e x^{e-1}+e^{x}=\frac{d t}{d x}$
$\Rightarrow\left(e x^{e-1}+e^{x}\right) d x=d t$
$\therefore \int \frac{\left(x^{e-1}+e^{x-1}\right) d x}{x^{e}+e^{x}}=\frac{1}{e} \int \frac{e x^{e-1}+e^{x}}{x^{e}+e^{x}} \cdot d x$
$=\frac{1}{e} \int \frac{d t}{t}=\frac{1}{e} \cdot \ln t+c$
$=\frac{1}{e} \cdot \ln \left(x^{e}+e^{x}\right)+c$
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