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: =∫1e1u−1u+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)

Best solutions to - "∫011ex+1dx"

tabita57i

Answered question

2021-09-15

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

Answer & Explanation

komunidadO

Skilled2021-09-16Added 86 answers

Compute the definite integral:
$\int_{0}^{1} \frac{1}{e^{x} + 1} d x$
For the integrand $\frac{1}{e^{x} + 1}$ , substitute $u = e^{x} and d u = e^{x} d x .$
This gives a new lower bound $u = e^{0} = 1$ and upper bound $u = e^{1} = e$ :
$= \int_{1}^{e} \frac{1}{u (u + 1)} d u$
For the integrand $\frac{1}{u (u + 1)}$ , 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 $s = u + 1 and d s = d u$ .
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):
$= (- \log (s)) |_{2}^{1 + e} + \int_{1}^{e} \frac{1}{u} d u$
Evaluate the antiderivative at the limits and subtract.
$(- \log (s)) |_{2}^{1 + e} = (- \log (1 + e)) - (- \log (2)) = \log (\frac{2}{1 + e})$ :
$= \log (\frac{2}{1 + e}) + \int_{1}^{e} \frac{1}{u} d u$
Apply the fundamental theorem of calculus.
The antiderivative of $\frac{1}{u}$ is $\log (u)$ :
$= \log (\frac{2}{1 + e}) + \log (u) |_{1}^{e}$
Evaluate the antiderivative at the limits and subtract.
$\log (u) |_{1}^{e} = \log (e) - \log (1) = 1 :$
$= 1 + \log (\frac{2}{1 + e})$
Which is equal to:
Answer:
$= 1 + \log (2) - \log (1 + e)$

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