Integral of $$$x^{2} \sqrt{x^{3}}$$$

Find $$$\int x^{2} \sqrt{x^{3}}\, dx$$$.

The input is rewritten: $$$\int{x^{2} \sqrt{x^{3}} d x}=\int{x^{\frac{7}{2}} d x}$$$.

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{7}{2}$$$:

$${\color{red}{\int{x^{\frac{7}{2}} d x}}}={\color{red}{\frac{x^{1 + \frac{7}{2}}}{1 + \frac{7}{2}}}}={\color{red}{\left(\frac{2 x^{\frac{9}{2}}}{9}\right)}}$$

Therefore,

$$\int{x^{\frac{7}{2}} d x} = \frac{2 x^{\frac{9}{2}}}{9}$$

Add the constant of integration:

$$\int{x^{\frac{7}{2}} d x} = \frac{2 x^{\frac{9}{2}}}{9}+C$$

$$$\int x^{2} \sqrt{x^{3}}\, dx = \frac{2 x^{\frac{9}{2}}}{9} + C$$$A