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Evaluate the integrals. Assume a, b, A, B, P0P_{0}, h, and k are constants. ex2+exdx\int \frac{e^{x}}{2+e^{x}} d x

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ex2+exdx=1uduSubstitute u=2+exdu=ex  dx=lnu+C=ln2+ex+Cu=2+ex=ln(2+ex)+C2+ex>0\begin{align*} \int \frac{e^x}{2+e^x} \, \text{d}x &= \int \frac{1}{u} \, \text{d}u & & \color{#5a140a} \begin{gathered} \text{Substitute }\\ u=2+e^x \Rightarrow \text{d}u = e^x \; \text{d} x \end{gathered} \\[4pt] & = \ln \left| u \right| + C \\[4pt] & = \ln \left| 2+e^x \right| + C & & \color{#5a140a} {u=2+e^x } \\[4pt] & = \ln \left( 2+e^x \right) +C & & \color{#5a140a} {2+e^x>0 } \end{align*}

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