Question

{\int_{{0}}^{{1}}}\frac{1}{{{e}^{x}+{1}}}{\left.{d}{x}\right.}

Integrals
ANSWERED
asked 2021-09-15

\({\int_{{0}}^{{1}}}\frac{1}{{{e}^{x}+{1}}}{\left.{d}{x}\right.}\)

Expert Answers (1)

2021-09-16

Compute the definite integral:
\({\int_{{0}}^{{1}}}\frac{1}{{{e}^{x}+{1}}}{\left.{d}{x}\right.}\)
For the integrand \(\frac{1}{{{e}^{x}+{1}}}\), substitute \({u}={e}^{x}{\quad\text{and}\quad}{d}{u}={e}^{x}{\left.{d}{x}\right.}.\)
This gives a new lower bound \({u}={e}^{0}={1}\) and upper bound \({u}={e}^{1}={e}\):
\(={\int_{{1}}^{{e}}}\frac{1}{{{u}{\left({u}+{1}\right)}}}{d}{u}\)
For the integrand \(\frac{1}{{{u}{\left({u}+{1}\right)}}}\), use partial fractions:
\(={\int_{{1}}^{{e}}}\frac{1}{{u}}-\frac{1}{{{u}+{1}}}{d}{u}\)
Integrate the sum term by term and factor out constants:
\(=-{\int_{{1}}^{{e}}}\frac{1}{{{u}+{1}}}{d}{u}+{\int_{{1}}^{{e}}}\frac{1}{{u}}{d}{u}\)
For the integrand \(\frac{1}{{{u}+{1}}}\), substitute \({s}={u}+{1}{\quad\text{and}\quad}{d}{s}={d}{u}\).
This gives a new lower bound \({s}={1}+{1}={2}\) and upper bound \({s}={1}+{e}\):
\(=-{\int_{{2}}^{{{1}+{e}}}}\frac{1}{{s}}{d}{s}+{\int_{{1}}^{{e}}}\frac{1}{{u}}{d}{u}\)
Apply the fundamental theorem of calculus.
The antiderivative of \(\frac{1}{{s}}\) is log(s):
\(={\left(- \log{{\left({s}\right)}}\right)}{{|}_{{2}}^{{{1}+{e}}}}+{\int_{{1}}^{{e}}}\frac{1}{{u}}{d}{u}\)
Evaluate the antiderivative at the limits and subtract.
\({\left(- \log{{\left({s}\right)}}\right)}{{|}_{{2}}^{{{1}+{e}}}}={\left(- \log{{\left({1}+{e}\right)}}\right)}-{\left(- \log{{\left({2}\right)}}\right)}= \log{{\left(\frac{2}{{{1}+{e}}}\right)}}\):
\(= \log{{\left(\frac{2}{{{1}+{e}}}\right)}}+{\int_{{1}}^{{e}}}\frac{1}{{u}}{d}{u}\)
Apply the fundamental theorem of calculus.
The antiderivative of \(\frac{1}{{u}}\) is \(\log{{\left({u}\right)}}\):
\(= \log{{\left(\frac{2}{{{1}+{e}}}\right)}}+ \log{{\left({u}\right)}}{{|}_{{1}}^{{e}}}\)
Evaluate the antiderivative at the limits and subtract.
\(\log{{\left({u}\right)}}{{|}_{{1}}^{{e}}}= \log{{\left({e}\right)}}- \log{{\left({1}\right)}}={1}:\)
\(={1}+ \log{{\left(\frac{2}{{{1}+{e}}}\right)}}\)
Which is equal to:
Answer:
\(={1}+ \log{{\left({2}\right)}}- \log{{\left({1}+{e}\right)}}\)

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