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

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

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

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

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

Decimal

1.59726 \dots

Solution steps

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

\int_{0}^{x} 3 x^{2} d x

\int_{1}^{e} y ln (y) d y

\int_{0}^{x^{2}} \frac{1}{x ^{3} + 1} d x

\int_{0}^{0} f (t) d t

\int_{0}^{x} t e^{- t^{2}} d t

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