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

Armorikam 2021-06-16 Answered

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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Expert Answer

Jaylen Fountain
Answered 2021-06-17 Author has 16342 answers

\(\displaystyle\int{\frac{{{1}}}{{{2}+{e}^{{-{x}}}}}}{\left.{d}{x}\right.}\)
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, \(\displaystyle{u}={2}{e}^{{{x}}}+{1}\)
\(\displaystyle{\frac{{{1}}}{{{2}}}}{\ln{{\left({2}{e}^{{{x}}}+{1}\right)}}}+{C}\)

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