# Evaluate the following integrals. int_{-2}^2frac{e^{z/2}}{e^{z/2}+1}dz

Evaluate the following integrals.
${\int }_{-2}^{2}\frac{{e}^{z/2}}{{e}^{z/2}+1}dz$
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To Determine: evaluate the following integral .include absolute values only when needed.

$\text{Explanation: we have}{\int }_{-2}^{2}\frac{{e}^{z/2}}{{e}^{z/2}+1}dz$

$u={e}^{z/2}+1$
$\text{then}$
$du=\frac{1}{2}{e}^{z/2}dz$
$2du={e}^{z/2}dz$

$\text{so our integral becomes}$
${\int }_{-2}^{2}\frac{{e}^{z/2}}{{e}^{z/2}+1}dz={\int }_{\frac{1}{e}+1}^{e+1}\frac{2}{u}du$
$=2\left[\mathrm{ln}|u|{\right]}_{\frac{1}{e}+1}^{e+1}$
$=2\left[\mathrm{ln}\left(e+1\right)-\mathrm{ln}\left(\frac{1}{e}+1\right)\right]$
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