What is the integral from 1 to 3 of 3r^4ln(r) ?

The integral from 1 to 3 of 3r^4ln(r) is 3(-242/25+243/5 ln(3))

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integral from 1 to 3 of 3r^4ln(r)

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

3 (- \frac{242}{25} + \frac{243}{5} ln (3))

Decimal Notation

131.13767 \dots

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{1}^{3} r^{4} ln (r) d r

Apply Integration By Parts

: [\frac{1}{5} r^{5} ln (r) - \int \frac{r ^{4}}{5} d r]_{1}^{3}

= 3 [\frac{1}{5} r^{5} ln (r) - \int \frac{r ^{4}}{5} d r]_{1}^{3}

\int \frac{r ^{4}}{5} d r = \frac{r ^{5}}{25}

= 3 [\frac{1}{5} r^{5} ln (r) - \frac{r ^{5}}{25}]_{1}^{3}

Compute the boundaries:

- \frac{242}{25} + \frac{243}{5} ln (3)

= 3 (- \frac{242}{25} + \frac{243}{5} ln (3))

d y - (y - 5)^{2} d x = 0

x^{2}, 2 x^{4}, - 1, 1

x \to 0 lim (\frac{1}{sin ^{2} ( x )})

\int_{0}^{1} 30 \frac{1}{x ^{p}} d x

\frac{4 x + x ^{2}}{( 2 + x ) ^{2}}

What is the integral from 1 to 3 of 3r^4ln(r) ?
The integral from 1 to 3 of 3r^4ln(r) is 3(-242/25+243/5 ln(3))