What is the integral of (r^3+1)^2r^2 ?

The integral of (r^3+1)^2r^2 is (r^9)/9+(r^6)/3+(r^3)/3+C

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integral of (r^3+1)^2r^2

\int (r^{3} + 1)^{2} r^{2} d r

\frac{r ^{9}}{9} + \frac{r ^{6}}{3} + \frac{r ^{3}}{3} + C

Solution steps

\int (r^{3} + 1)^{2} r^{2} d r

Expand

(r^{3} + 1)^{2} r^{2} : r^{8} + 2 r^{5} + r^{2}

= \int r^{8} + 2 r^{5} + r^{2} 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 r^{8} d r + \int 2 r^{5} d r + \int r^{2} d r

\int r^{8} d r = \frac{r ^{9}}{9}

\int 2 r^{5} d r = \frac{r ^{6}}{3}

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

= \frac{r ^{9}}{9} + \frac{r ^{6}}{3} + \frac{r ^{3}}{3}

Add a constant to the solution

= \frac{r ^{9}}{9} + \frac{r ^{6}}{3} + \frac{r ^{3}}{3} + C

\frac{1 - ln ( x )}{x ^{2}}

\int_{4}^{15} 6 x^{- 2} d x

4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 2 x

\frac{d}{d s} (\frac{s + 2}{s ^{2} + 4 s + 13})

f (x) = x^{2} + 6, (5, 31)

What is the integral of (r^3+1)^2r^2 ?
The integral of (r^3+1)^2r^2 is (r^9)/9+(r^6)/3+(r^3)/3+C