What is the integral from 0 to 2 of xln(x+1) ?

The integral from 0 to 2 of xln(x+1) is 3/2 ln(3)

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integral from 0 to 2 of xln(x+1)

\int_{0}^{2} x ln (x + 1) d x

\frac{3}{2} ln (3)

Decimal

1.64791 \dots

Solution steps

\int_{0}^{2} x ln (x + 1) d x

Apply u-substitution

= \int_{1}^{3} (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}^{3}

\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}^{3}

Simplify

[ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3} : [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

= [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

Compute the boundaries:

\frac{3}{2} ln (3)

= \frac{3}{2} ln (3)

What is the integral from 0 to 2 of xln(x+1) ?
The integral from 0 to 2 of xln(x+1) is 3/2 ln(3)