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

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

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

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

\int \frac{x ^{4} - x ^{3}}{x ^{5}} d x

\frac{d}{d x} (\frac{x ^{3} + 9}{x})

y^{^{'}} = \frac{y ^{3} + 1}{( 3 x y ^{2} )}

\frac{\partial}{\partial x} (3 x y^{2})

\frac{d}{d x} (4 x^{3} y^{3})

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