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 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} (y e^{- x})

\frac{d P}{d t} = a P - b P^{2}

x \to 2 lim (\frac{( x + 3 )}{x - 2})

\frac{d y}{d x} = \frac{e ^{x}}{6 + e ^{x}}

cos (x), 0.5, 0 \leq x \leq π

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)