What is the integral from 0 to 1 of e^xln(e^x+1) ?

The integral from 0 to 1 of e^xln(e^x+1) is eln(e+1)+ln(e+1)+1-2ln(2)-e

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integral from 0 to 1 of e^xln(e^x+1)

\int_{0}^{1} e^{x} ln (e^{x} + 1) d x

e ln (e + 1) + ln (e + 1) + 1 - 2 ln (2) - e

Solution steps

\int_{0}^{1} e^{x} ln (e^{x} + 1) d x

Apply u-substitution

: \int_{2}^{e + 1} ln (u) d u

= \int_{2}^{e + 1} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{2}^{e + 1}

= [u ln (u) - \int 1 d u]_{2}^{e + 1}

\int 1 d u = u

= [u ln (u) - u]_{2}^{e + 1}

Compute the boundaries:

e ln (e + 1) + ln (e + 1) + 1 - 2 ln (2) - e

= e ln (e + 1) + ln (e + 1) + 1 - 2 ln (2) - e

\int_{0}^{3} \frac{x}{1 + x ^{2}} d x

\int_{2}^{7} 700 x 3 + 2 x d x

\int_{0}^{1} (2 - e^{x})^{2} d x

\int_{3}^{1} 1

\int_{0}^{4} (11 - x) x d x

What is the integral from 0 to 1 of e^xln(e^x+1) ?
The integral from 0 to 1 of e^xln(e^x+1) is eln(e+1)+ln(e+1)+1-2ln(2)-e