What is the integral from 1 to e^4 of sqrt(x)ln(x) ?

The integral from 1 to e^4 of sqrt(x)ln(x) is (20e^6+4)/9

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integral from 1 to e^4 of sqrt(x)ln(x)

\int_{1}^{e^{4}} x ln (x) d x

\frac{20 e ^{6} + 4}{9}

Decimal

896.95287 \dots

Solution steps

\int_{1}^{e^{4}} 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^{4}}

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

\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^{4}}

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^{4}}

Compute the boundaries:

\frac{20 e ^{6}}{9} + \frac{4}{9}

= \frac{20 e ^{6}}{9} + \frac{4}{9}

Simplify

= \frac{20 e ^{6} + 4}{9}

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What is the integral from 1 to e^4 of sqrt(x)ln(x) ?
The integral from 1 to e^4 of sqrt(x)ln(x) is (20e^6+4)/9