Integral of $$$\frac{x^{2}}{\sqrt{x^{21}}}$$$

Find $$$\int \frac{x^{2}}{\sqrt{x^{21}}}\, dx$$$.

The input is rewritten: $$$\int{\frac{x^{2}}{\sqrt{x^{21}}} d x}=\int{\frac{1}{x^{\frac{17}{2}}} d x}$$$.

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{17}{2}$$$:

$${\color{red}{\int{\frac{1}{x^{\frac{17}{2}}} d x}}}={\color{red}{\int{x^{- \frac{17}{2}} d x}}}={\color{red}{\frac{x^{- \frac{17}{2} + 1}}{- \frac{17}{2} + 1}}}={\color{red}{\left(- \frac{2 x^{- \frac{15}{2}}}{15}\right)}}={\color{red}{\left(- \frac{2}{15 x^{\frac{15}{2}}}\right)}}$$

Therefore,

$$\int{\frac{1}{x^{\frac{17}{2}}} d x} = - \frac{2}{15 x^{\frac{15}{2}}}$$

Add the constant of integration:

$$\int{\frac{1}{x^{\frac{17}{2}}} d x} = - \frac{2}{15 x^{\frac{15}{2}}}+C$$

$$$\int \frac{x^{2}}{\sqrt{x^{21}}}\, dx = - \frac{2}{15 x^{\frac{15}{2}}} + C$$$A