What is the integral from 1 to 5 of sqrt(x)ln(x) ?

The integral from 1 to 5 of sqrt(x)ln(x) is (10sqrt(5))/3 ln(5)+(-20sqrt(5)+4)/9

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integral from 1 to 5 of sqrt(x)ln(x)

\int_{1}^{5} x ln (x) d x

\frac{10 5}{3} ln (5) + \frac{- 20 5 + 4}{9}

Decimal Notation

7.47144 \dots

Solution steps

\int_{1}^{5} 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}^{5}

= [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \int \frac{2}{3} x^{\frac{1}{2}} d x]_{1}^{5}

\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}^{5}

Compute the boundaries:

\frac{10 5}{3} ln (5) - \frac{20 5}{9} + \frac{4}{9}

= \frac{10 5}{3} ln (5) - \frac{20 5}{9} + \frac{4}{9}

Simplify

= \frac{10 5}{3} ln (5) + \frac{- 20 5 + 4}{9}

f (x) = x (x + 3)^{2}

\int_{4}^{\infty} \frac{1}{x ^{3}} d x

y^{^{'}} - 2 y = 7

\frac{\partial}{\partial x} (x + yz)

\frac{d}{d x} (6 + x)

What is the integral from 1 to 5 of sqrt(x)ln(x) ?
The integral from 1 to 5 of sqrt(x)ln(x) is (10sqrt(5))/3 ln(5)+(-20sqrt(5)+4)/9