What is the integral from 0 to r of (br^2+cr) ?

The integral from 0 to r of (br^2+cr) is (br^3)/3+(cr^2)/2

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integral from 0 to r of (br^2+cr)

\int_{0}^{r} (b r^{2} + cr) d r

\frac{b r ^{3}}{3} + \frac{c r ^{2}}{2}

Solution steps

\int_{0}^{r} b r^{2} + cr d r

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{r} b r^{2} d r + \int_{0}^{r} cr d r

\int_{0}^{r} b r^{2} d r = \frac{b r ^{3}}{3}

\int_{0}^{r} cr d r = \frac{c r ^{2}}{2}

= \frac{b r ^{3}}{3} + \frac{c r ^{2}}{2}

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

What is the integral from 0 to r of (br^2+cr) ?
The integral from 0 to r of (br^2+cr) is (br^3)/3+(cr^2)/2