To Determine: evaluate the following integral .include absolute values only when needed.

\(\text{Given: we have an integral }\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)

\(\text{Explanation: we have}\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)

\(\text{let us consider }\)

\(u=e^{z/2}+1\)

\(\text{then}\)

\(du=\frac{1}{2}e^{z/2}dz\)

\(2du=e^{z/2}dz\)

\(\text{the limits also get changed }\)

\(z=-2\ then\ u=e^{-2/2}+1\ =e^{-1}+1\ =\frac{1}{e}+1\)

\(z=2\ then\ u=e^{2/2}+1\ =e^1+1\)

\(\text{so our integral becomes}\)

\(\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz=\int_{\frac{1}{e}+1}^{e+1}\frac{2}{u}du\)

\(=2[\ln|u|]_{\frac{1}{e}+1}^{e+1}\)

\(=2[\ln(e+1)-\ln(\frac{1}{e}+1)]\)

\(\text{Given: we have an integral }\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)

\(\text{Explanation: we have}\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)

\(\text{let us consider }\)

\(u=e^{z/2}+1\)

\(\text{then}\)

\(du=\frac{1}{2}e^{z/2}dz\)

\(2du=e^{z/2}dz\)

\(\text{the limits also get changed }\)

\(z=-2\ then\ u=e^{-2/2}+1\ =e^{-1}+1\ =\frac{1}{e}+1\)

\(z=2\ then\ u=e^{2/2}+1\ =e^1+1\)

\(\text{so our integral becomes}\)

\(\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz=\int_{\frac{1}{e}+1}^{e+1}\frac{2}{u}du\)

\(=2[\ln|u|]_{\frac{1}{e}+1}^{e+1}\)

\(=2[\ln(e+1)-\ln(\frac{1}{e}+1)]\)