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$\int \frac{\mathrm{dx}}{1+\mathrm{e}^{-\mathrm{x}}}$ is equal to
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Verified Answer
The correct answer is:
$\ln \left(1+\mathrm{e}^{x}\right)+\mathrm{c}$
$\int \frac{\mathrm{dx}}{1+\mathrm{e}^{-\mathrm{x}}}$
$\Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx}$
Let $\mathrm{e}^{\mathrm{x}}+1=\mathrm{t}$
$\mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$=\int \frac{\mathrm{dt}}{\mathrm{t}}$
$\Rightarrow \log \mathrm{t}+\mathrm{c} \Rightarrow \log \left(\mathrm{e}^{\mathrm{x}}+1\right)+\mathrm{c}$
$\Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx}$
Let $\mathrm{e}^{\mathrm{x}}+1=\mathrm{t}$
$\mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$=\int \frac{\mathrm{dt}}{\mathrm{t}}$
$\Rightarrow \log \mathrm{t}+\mathrm{c} \Rightarrow \log \left(\mathrm{e}^{\mathrm{x}}+1\right)+\mathrm{c}$
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