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

The integral from 1 to e of (xln(x))^2 is (5e^3-2)/(27)

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

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

\frac{5 e ^{3} - 2}{27}

Decimal

3.64546 \dots

Solution steps

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

Apply Integration By Parts

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

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

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

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

Simplify

[\frac{1}{3} (x^{3} ln^{2} (x) - 3 \cdot \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}))]_{1}^{e} : [\frac{1}{3} (x^{3} ln^{2} (x) - 2 (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}))]_{1}^{e}

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

Compute the boundaries:

\frac{5 e ^{3}}{27} - \frac{2}{27}

= \frac{5 e ^{3}}{27} - \frac{2}{27}

Simplify

= \frac{5 e ^{3} - 2}{27}

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What is the integral from 1 to e of (xln(x))^2 ?
The integral from 1 to e of (xln(x))^2 is (5e^3-2)/(27)