What is the integral from 1 to 4 of 3sqrt(x)ln(x) ?

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

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integral from 1 to 4 of 3sqrt(x)ln(x)

\int_{1}^{4} 3 x ln (x) d x

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

Decimal Notation

12.84737 \dots

Solution steps

\int_{1}^{4} 3 x ln (x) d x

Take the constant out:

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

= 3 \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}

= 3 [\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}}

= 3 [\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)

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

Simplify

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

\frac{d}{d x} (\frac{( x - 1 ) ^{4}}{( x ^{2} + 2 x ) ^{5}})

\frac{d}{d y} (e^{x} (x - y)^{2})

\frac{\partial}{\partial x} (cos^{3} (x))

\int (4 x^{2} - 2) d x

\frac{\partial}{\partial y} (\frac{4 x}{3 y ^{2} + 1})

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