What is the integral from 1 to e of 6x^2ln(x) ?

The integral from 1 to e of 6x^2ln(x) is (2(2e^3+1))/3

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integral from 1 to e of 6x^2ln(x)

\int_{1}^{e} 6 x^{2} ln (x) d x

\frac{2 ( 2 e ^{3} + 1 )}{3}

Decimal Notation

27.44738 \dots

Solution steps

\int_{1}^{e} 6 x^{2} ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \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}

= 6 [\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}

= 6 [\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}

= 6 [\frac{1}{3} (x^{3} ln (x) - \frac{x ^{3}}{3})]_{1}^{e}

Compute the boundaries:

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

= 6 (\frac{2 e ^{3}}{9} + \frac{1}{9})

Simplify

= \frac{2 ( 2 e ^{3} + 1 )}{3}

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What is the integral from 1 to e of 6x^2ln(x) ?
The integral from 1 to e of 6x^2ln(x) is (2(2e^3+1))/3