What is the integral from 0 to (5pi)/4 of (ln(x)x^2) ?

The integral from 0 to (5pi)/4 of (ln(x)x^2) is 1/3 ((125pi^3}{64}ln((5pi)/4)-\frac{125pi^3)/(192))

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integral from 0 to (5pi)/4 of (ln(x)x^2)

\int_{0}^{\frac{5 π}{4}} (ln (x) x^{2}) d x

\frac{1}{3} (\frac{125 π ^{3}}{64} ln (\frac{5 π}{4}) - \frac{125 π ^{3}}{192})

Decimal Notation

20.88361 \dots

Solution steps

\int_{0}^{\frac{5 π}{4}} ln (x) x^{2} d x

Apply Integration By Parts

: [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \int \frac{x ^{2}}{3} d x)]_{0}^{\frac{5 π}{4}}

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \int \frac{x ^{2}}{3} d x)]_{0}^{\frac{5 π}{4}}

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \frac{x ^{3}}{9})]_{0}^{\frac{5 π}{4}}

3 \cdot \frac{x ^{3}}{9} = \frac{x ^{3}}{3}

= [\frac{1}{3} (x^{3} ln (x) - \frac{x ^{3}}{3})]_{0}^{\frac{5 π}{4}}

Compute the boundaries:

\frac{1}{3} (\frac{125 π ^{3}}{64} ln (\frac{5 π}{4}) - \frac{125 π ^{3}}{192})

= \frac{1}{3} (\frac{125 π ^{3}}{64} ln (\frac{5 π}{4}) - \frac{125 π ^{3}}{192})

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What is the integral from 0 to (5pi)/4 of (ln(x)x^2) ?
The integral from 0 to (5pi)/4 of (ln(x)x^2) is 1/3 ((125pi^3}{64}ln((5pi)/4)-\frac{125pi^3)/(192))