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

The integral from 1 to e of x^2ln(x) is (2e^3+1)/9

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

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

\frac{2 e ^{3} + 1}{9}

Decimal Notation

4.57456 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

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

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

Compute the boundaries:

\frac{2 e ^{3}}{9} + \frac{1}{9}

= \frac{2 e ^{3}}{9} + \frac{1}{9}

Simplify

= \frac{2 e ^{3} + 1}{9}

\frac{d y}{d x} = (x y^{2})

x \to 0 lim (\frac{3 x}{tan ( x )})

y^{^{'}} - 2 x y = 2 x

x \to 1 lim (arctan (x))

\frac{\partial}{\partial x} (x sin (x y))

What is the integral from 1 to e of x^2ln(x) ?
The integral from 1 to e of x^2ln(x) is (2e^3+1)/9