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

The integral from 0 to e-1 of xln(x+1) is (e^2-3)/4

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

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

\frac{e ^{2} - 3}{4}

Decimal Notation

1.09726 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{e}

Simplify

[ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{e} : [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{1}^{e}

= [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{1}^{e}

Compute the boundaries:

\frac{e ^{2}}{4} - \frac{3}{4}

= \frac{e ^{2}}{4} - \frac{3}{4}

Simplify

= \frac{e ^{2} - 3}{4}

\int_{1}^{2} p^{2} ln (p) d p

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\int_{8}^{10} \frac{x}{8} - \frac{8}{x} d x

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

What is the integral from 0 to e-1 of xln(x+1) ?
The integral from 0 to e-1 of xln(x+1) is (e^2-3)/4