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

The integral from 0 to 1 of x^2(ln(1/x))^3 is 2/27

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integral from 0 to 1 of x^2(ln(1/x))^3

\int_{0}^{1} x^{2} (ln (\frac{1}{x}))^{3} d x

\frac{2}{27}

Solution steps

\int_{0}^{1} x^{2} (ln (\frac{1}{x}))^{3} d x

Simplify

x^{2} (ln (\frac{1}{x}))^{3} : - x^{2} ln^{3} (x)

= \int_{0}^{1} - x^{2} ln^{3} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{0}^{1} x^{2} ln^{3} (x) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} ln^{3} (x) - \int x^{2} ln^{2} (x) d x]_{0}^{1}

= - [\frac{1}{3} x^{3} ln^{3} (x) - \int x^{2} ln^{2} (x) d x]_{0}^{1}

\int x^{2} ln^{2} (x) d x = \frac{1}{3} x^{3} ln^{2} (x) - \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9})

= - [\frac{1}{3} x^{3} ln^{3} (x) - (\frac{1}{3} x^{3} ln^{2} (x) - \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}))]_{0}^{1}

Expand

\frac{1}{3} x^{3} ln^{3} (x) - (\frac{1}{3} x^{3} ln^{2} (x) - \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9})) : \frac{1}{3} x^{3} ln^{3} (x) - \frac{1}{3} x^{3} ln^{2} (x) + \frac{2}{9} x^{3} ln (x) - \frac{2 x ^{3}}{27}

= - [\frac{1}{3} x^{3} ln^{3} (x) - \frac{1}{3} x^{3} ln^{2} (x) + \frac{2}{9} x^{3} ln (x) - \frac{2 x ^{3}}{27}]_{0}^{1}

Compute the boundaries:

- \frac{2}{27}

= - (- \frac{2}{27})

Simplify

= \frac{2}{27}

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What is the integral from 0 to 1 of x^2(ln(1/x))^3 ?
The integral from 0 to 1 of x^2(ln(1/x))^3 is 2/27