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

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

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

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

e ln (2) + \frac{- Ei ( 1 + 1 ) + Ei ( 0 + 1 )}{e}

Solution steps

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

Apply Integration By Parts

: [e^{x} ln (x + 1) - \int \frac{e ^{x}}{x + 1} d x]_{0}^{1}

= [e^{x} ln (x + 1) - \int \frac{e ^{x}}{x + 1} d x]_{0}^{1}

\int \frac{e ^{x}}{x + 1} d x = \frac{1}{e} Ei (x + 1)

= [e^{x} ln (x + 1) - \frac{1}{e} Ei (x + 1)]_{0}^{1}

Compute the boundaries:

e ln (2) + \frac{- Ei ( 1 + 1 ) + Ei ( 0 + 1 )}{e}

= e ln (2) + \frac{- Ei ( 1 + 1 ) + Ei ( 0 + 1 )}{e}

\int_{- 2}^{0} \frac{1}{x} d x

\int_{2}^{4} \frac{2}{x} d x

\int_{- 1}^{1} ∣ x - 2 ∣ d x

\int_{0}^{2} (6 - t) t d t

\int_{0}^{\infty} e^{- n x} d x

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