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

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

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

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

\frac{1}{4}

Solution steps

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

Apply Integration By Parts

: [\frac{1}{2} x^{2} ln^{2} (x) - \int x ln (x) d x]_{0}^{1}

= [\frac{1}{2} x^{2} ln^{2} (x) - \int x ln (x) d x]_{0}^{1}

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

= [\frac{1}{2} x^{2} ln^{2} (x) - (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})]_{0}^{1}

Simplify

= [\frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= \frac{1}{4}

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