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

The integral from 1 to 4 of sqrt(x)*ln(x) is -28/9+32/3 ln(2)

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integral from 1 to 4 of sqrt(x)*ln(x)

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

- \frac{28}{9} + \frac{32}{3} ln (2)

Decimal Notation

4.28245 \dots

Solution steps

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

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

\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}^{4}

Compute the boundaries:

- \frac{28}{9} + \frac{32}{3} ln (2)

= - \frac{28}{9} + \frac{32}{3} ln (2)

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{8}{3} - x^{2} + 4)

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'}} = 6 x

\frac{d}{d x} (- x^{2} + 2 x + 1)

\frac{\partial}{\partial y} (e^{x + 2 y})

What is the integral from 1 to 4 of sqrt(x)*ln(x) ?
The integral from 1 to 4 of sqrt(x)*ln(x) is -28/9+32/3 ln(2)