What is the integral from 1 to e of x^4(ln(x))^2 ?

The integral from 1 to e of x^4(ln(x))^2 is (17e^5-2)/(125)

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integral from 1 to e of x^4(ln(x))^2

\int_{1}^{e} x^{4} (ln (x))^{2} d x

\frac{17 e ^{5} - 2}{125}

Solution steps

\int_{1}^{e} x^{4} (ln (x))^{2} d x

Apply Integration By Parts

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

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

\int \frac{2 x ^{4} ln ( x )}{5} d x = \frac{2}{5} (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25})

= [\frac{1}{5} (x^{5} ln^{2} (x) - 5 \cdot \frac{2}{5} (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}))]_{1}^{e}

Simplify

[\frac{1}{5} (x^{5} ln^{2} (x) - 5 \cdot \frac{2}{5} (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}))]_{1}^{e} : [\frac{1}{5} (x^{5} ln^{2} (x) - 2 (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}))]_{1}^{e}

= [\frac{1}{5} (x^{5} ln^{2} (x) - 2 (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}))]_{1}^{e}

Compute the boundaries:

\frac{17 e ^{5}}{125} - \frac{2}{125}

= \frac{17 e ^{5}}{125} - \frac{2}{125}

Simplify

= \frac{17 e ^{5} - 2}{125}

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What is the integral from 1 to e of x^4(ln(x))^2 ?
The integral from 1 to e of x^4(ln(x))^2 is (17e^5-2)/(125)