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

Question
Integrals
Evaluate the following integrals.
$$\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz$$

2021-01-09
To Determine: evaluate the following integral .include absolute values only when needed.
$$\text{Given: we have an integral }\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz$$
$$\text{Explanation: we have}\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz$$
$$\text{let us consider }$$
$$u=e^{z/2}+1$$
$$\text{then}$$
$$du=\frac{1}{2}e^{z/2}dz$$
$$2du=e^{z/2}dz$$
$$\text{the limits also get changed }$$
$$z=-2\ then\ u=e^{-2/2}+1\ =e^{-1}+1\ =\frac{1}{e}+1$$
$$z=2\ then\ u=e^{2/2}+1\ =e^1+1$$
$$\text{so our integral becomes}$$
$$\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz=\int_{\frac{1}{e}+1}^{e+1}\frac{2}{u}du$$
$$=2[\ln|u|]_{\frac{1}{e}+1}^{e+1}$$
$$=2[\ln(e+1)-\ln(\frac{1}{e}+1)]$$

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