# Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \int \frac{1}{2+e^

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.
$$\displaystyle\int{\frac{{{1}}}{{{2}+{e}^{{-{x}}}}}}{\left.{d}{x}\right.}$$

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$$\displaystyle\int{\frac{{{1}}}{{{2}+{e}^{{-{x}}}}}}{\left.{d}{x}\right.}$$
Apply the negative exponent rule $$\frac{1}{a^{n}}=a^{-n}$$
$$\displaystyle\int{\frac{{{1}}}{{{2}+{e}^{{-{x}}}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{{2}+{\frac{{{1}}}{{{e}^{{{x}}}}}}}}}{\left.{d}{x}\right.}=\int{\frac{{{e}^{{{x}}}}}{{{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}$$
Let $$u=2e^{x}\Rightarrow du=2e^{x}dx\Rightarrow \frac{1}{2}du=e^{x}dx$$
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}=\int{\frac{{{\frac{{{1}}}{{{2}}}}{d}{u}}}{{{u}}}}={\frac{{{1}}}{{{2}}}}\int{\frac{{{d}{u}}}{{{u}}}}$$
Integrate
$$\displaystyle{\frac{{{1}}}{{{2}}}}\int{\frac{{{d}{u}}}{{{u}}}}={\frac{{{1}}}{{{2}}}}{\ln}{\left|{u}\right|}+{C}$$
Putting the answer back in terms of x, $$\displaystyle{u}={2}{e}^{{{x}}}+{1}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{\ln{{\left({2}{e}^{{{x}}}+{1}\right)}}}+{C}$$