What is the integral from 1 to 2 of 4r^2ln(r) ?

The integral from 1 to 2 of 4r^2ln(r) is 4(-7/9+8/3 ln(2))

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integral from 1 to 2 of 4r^2ln(r)

\int_{1}^{2} 4 r^{2} ln (r) d r

4 (- \frac{7}{9} + \frac{8}{3} ln (2))

Decimal Notation

4.28245 \dots

Solution steps

\int_{1}^{2} 4 r^{2} ln (r) d r

Take the constant out:

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

= 4 \cdot \int_{1}^{2} r^{2} ln (r) d r

Apply Integration By Parts

: [\frac{1}{3} r^{3} ln (r) - \int \frac{r ^{2}}{3} d r]_{1}^{2}

= 4 [\frac{1}{3} r^{3} ln (r) - \int \frac{r ^{2}}{3} d r]_{1}^{2}

\int \frac{r ^{2}}{3} d r = \frac{r ^{3}}{9}

= 4 [\frac{1}{3} r^{3} ln (r) - \frac{r ^{3}}{9}]_{1}^{2}

Compute the boundaries:

- \frac{7}{9} + \frac{8}{3} ln (2)

= 4 (- \frac{7}{9} + \frac{8}{3} ln (2))

\frac{d}{d x} (ln (\frac{x - 2}{x + 1}))

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

\frac{11}{3} t^{3} + 11 t^{2}

\frac{d}{d x} ((6 x^{2} + 6) e^{- x})

What is the integral from 1 to 2 of 4r^2ln(r) ?
The integral from 1 to 2 of 4r^2ln(r) is 4(-7/9+8/3 ln(2))