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

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

The input is rewritten: $$$\int{\sqrt{x \sqrt{x^{\frac{3}{2}}}} d x}=\int{x^{\frac{7}{8}} d x}$$$.

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{7}{8}$$$:

$${\color{red}{\int{x^{\frac{7}{8}} d x}}}={\color{red}{\frac{x^{\frac{7}{8} + 1}}{\frac{7}{8} + 1}}}={\color{red}{\left(\frac{8 x^{\frac{15}{8}}}{15}\right)}}$$

Therefore,

$$\int{x^{\frac{7}{8}} d x} = \frac{8 x^{\frac{15}{8}}}{15}$$

Add the constant of integration:

$$\int{x^{\frac{7}{8}} d x} = \frac{8 x^{\frac{15}{8}}}{15}+C$$

$$$\int \sqrt{x \sqrt{x^{\frac{3}{2}}}}\, dx = \frac{8 x^{\frac{15}{8}}}{15} + C$$$A