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

tabita57i

tabita57i

Answered question

2021-09-15

011ex+1dx

Answer & Explanation

komunidadO

komunidadO

Skilled2021-09-16Added 86 answers

Compute the definite integral:
011ex+1dx
For the integrand 1ex+1, substitute u=exanddu=exdx.
This gives a new lower bound u=e0=1 and upper bound u=e1=e:
=1e1u(u+1)du
For the integrand 1u(u+1), use partial fractions:
=1e1u1u+1du
Integrate the sum term by term and factor out constants:
=1e1u+1du+1e1udu
For the integrand 1u+1, substitute s=u+1andds=du.
This gives a new lower bound s=1+1=2 and upper bound s=1+e:
=21+e1sds+1e1udu
Apply the fundamental theorem of calculus.
The antiderivative of 1s is log(s):
=(log(s))|21+e+1e1udu
Evaluate the antiderivative at the limits and subtract.
(log(s))|21+e=(log(1+e))(log(2))=log(21+e):
=log(21+e)+1e1udu
Apply the fundamental theorem of calculus.
The antiderivative of 1u is log(u):
=log(21+e)+log(u)|1e
Evaluate the antiderivative at the limits and subtract.
log(u)|1e=log(e)log(1)=1:
=1+log(21+e)
Which is equal to:
Answer:
=1+log(2)log(1+e)

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