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

The integral from 0 to 1 of x^2ln(x+1) is -5/18+2/3 ln(2)

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integral from 0 to 1 of x^2ln(x+1)

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

- \frac{5}{18} + \frac{2}{3} ln (2)

Decimal

0.18432 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int \frac{1}{3} (u^{2} - 3 u + 3) d u = \frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u

= [ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)]_{1}^{2}

Simplify

[ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)]_{1}^{2} : [\frac{1}{3} u^{3} ln (u) - u^{2} ln (u) + u ln (u) - \frac{u ^{3}}{9} + \frac{u ^{2}}{2} - u]_{1}^{2}

= [\frac{1}{3} u^{3} ln (u) - u^{2} ln (u) + u ln (u) - \frac{u ^{3}}{9} + \frac{u ^{2}}{2} - u]_{1}^{2}

Compute the boundaries:

- \frac{5}{18} + \frac{2}{3} ln (2)

= - \frac{5}{18} + \frac{2}{3} ln (2)

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What is the integral from 0 to 1 of x^2ln(x+1) ?
The integral from 0 to 1 of x^2ln(x+1) is -5/18+2/3 ln(2)