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

The integral from 1 to 2 of sqrt(x)ln(x) is (4sqrt(2))/3 ln(2)+(-8sqrt(2)+4)/9

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integral from 1 to 2 of sqrt(x)ln(x)

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

\frac{4 2}{3} ln (2) + \frac{- 8 2 + 4}{9}

Decimal Notation

0.49437 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

\frac{4 2}{3} ln (2) - \frac{8 2}{9} + \frac{4}{9}

= \frac{4 2}{3} ln (2) - \frac{8 2}{9} + \frac{4}{9}

Simplify

= \frac{4 2}{3} ln (2) + \frac{- 8 2 + 4}{9}

\frac{d}{d x} (e^{(- 3 x)})

e^{- 3 z + 1} + 8

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 5, (1, 8)

f (x) = \frac{2 x - 1}{2 x + 1}

t e^{- t}

What is the integral from 1 to 2 of sqrt(x)ln(x) ?
The integral from 1 to 2 of sqrt(x)ln(x) is (4sqrt(2))/3 ln(2)+(-8sqrt(2)+4)/9