What is the integral from 1 to 4 of 8sqrt(x)ln(x) ?

The integral from 1 to 4 of 8sqrt(x)ln(x) is 8(-28/9+32/3 ln(2))

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integral from 1 to 4 of 8sqrt(x)ln(x)

\int_{1}^{4} 8 x ln (x) d x

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

Decimal Notation

34.25967 \dots

Solution steps

\int_{1}^{4} 8 x ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \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}

= 8 [\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}}

= 8 [\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)

= 8 (- \frac{28}{9} + \frac{32}{3} ln (2))

\int (x^{3} + 4 x y) d x

\int_{0}^{1} (x^{2} + 1) (e^{- x}) d x

\frac{1}{s ^{2} + 4 s + 8}

g (x) = \frac{3}{6 x + 8}

\int x^{2} - 2 x - 3 d x

What is the integral from 1 to 4 of 8sqrt(x)ln(x) ?
The integral from 1 to 4 of 8sqrt(x)ln(x) is 8(-28/9+32/3 ln(2))