Question

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

Integrals

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

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:
$$={1}+ \log{{\left({2}\right)}}- \log{{\left({1}+{e}\right)}}$$