Integral of $$$\sqrt{x^{21}}$$$

Find $$$\int \sqrt{x^{21}}\, dx$$$.

The input is rewritten: $$$\int{\sqrt{x^{21}} d x}=\int{x^{\frac{21}{2}} d x}$$$.

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{21}{2}$$$:

$${\color{red}{\int{x^{\frac{21}{2}} d x}}}={\color{red}{\frac{x^{1 + \frac{21}{2}}}{1 + \frac{21}{2}}}}={\color{red}{\left(\frac{2 x^{\frac{23}{2}}}{23}\right)}}$$

Therefore,

$$\int{x^{\frac{21}{2}} d x} = \frac{2 x^{\frac{23}{2}}}{23}$$

Add the constant of integration:

$$\int{x^{\frac{21}{2}} d x} = \frac{2 x^{\frac{23}{2}}}{23}+C$$

$$$\int \sqrt{x^{21}}\, dx = \frac{2 x^{\frac{23}{2}}}{23} + C$$$A