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

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

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

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

\frac{e ^{2} - 1}{4}

Decimal Notation

1.59726 \dots

Solution steps

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

Apply Integration By Parts

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

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

\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) - 2 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}))]_{1}^{e}

Simplify

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

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

Compute the boundaries:

\frac{e ^{2}}{4} - \frac{1}{4}

= \frac{e ^{2}}{4} - \frac{1}{4}

Simplify

= \frac{e ^{2} - 1}{4}

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\frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}

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\int (1 - cos^{2} (x)) d x

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