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

The integral from 1 to e of sqrt(x)ln(x) is (2e^{3/2}+4)/9

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integral from 1 to e of sqrt(x)ln(x)

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

\frac{2 e ^{\frac{3}{2}} + 4}{9}

Decimal

1.44037 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 x ^{\frac{1}{2}}}{3} d x = \frac{4}{9} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - 3 \cdot \frac{4}{9} x^{\frac{3}{2}})]_{1}^{e}

3 \cdot \frac{4}{9} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - \frac{4}{3} x^{\frac{3}{2}})]_{1}^{e}

Compute the boundaries:

\frac{2 e ^{\frac{3}{2}}}{9} + \frac{4}{9}

= \frac{2 e ^{\frac{3}{2}}}{9} + \frac{4}{9}

Simplify

= \frac{2 e ^{\frac{3}{2}} + 4}{9}

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\int ln (6 x^{2}) d x

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What is the integral from 1 to e of sqrt(x)ln(x) ?
The integral from 1 to e of sqrt(x)ln(x) is (2e^{3/2}+4)/9