What is the integral from 1 to 3 of 9r^3ln(r) ?

The integral from 1 to 3 of 9r^3ln(r) is 9(-5+81/4 ln(3))

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integral from 1 to 3 of 9r^3ln(r)

\int_{1}^{3} 9 r^{3} ln (r) d r

9 (- 5 + \frac{81}{4} ln (3))

Decimal Notation

155.22208 \dots

Solution steps

\int_{1}^{3} 9 r^{3} ln (r) d r

Take the constant out:

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

= 9 \cdot \int_{1}^{3} r^{3} ln (r) d r

Apply Integration By Parts

: [\frac{1}{4} r^{4} ln (r) - \int \frac{r ^{3}}{4} d r]_{1}^{3}

= 9 [\frac{1}{4} r^{4} ln (r) - \int \frac{r ^{3}}{4} d r]_{1}^{3}

\int \frac{r ^{3}}{4} d r = \frac{r ^{4}}{16}

= 9 [\frac{1}{4} r^{4} ln (r) - \frac{r ^{4}}{16}]_{1}^{3}

Compute the boundaries:

- 5 + \frac{81}{4} ln (3)

= 9 (- 5 + \frac{81}{4} ln (3))

(sec^{3} (x))^{^{'}}

\int sin (- 7 x) d x

\int_{0}^{2} \frac{x ^{3}}{2} d x

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

y^{^{'}} = \frac{x y ( 4 - y )}{1 + x}

What is the integral from 1 to 3 of 9r^3ln(r) ?
The integral from 1 to 3 of 9r^3ln(r) is 9(-5+81/4 ln(3))