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

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

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

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

Therefore,

$$\int{x^{- \sqrt{2}} d x} = \frac{1}{x^{-1 + \sqrt{2}} \left(1 - \sqrt{2}\right)}$$

Add the constant of integration:

$$\int{x^{- \sqrt{2}} d x} = \frac{1}{x^{-1 + \sqrt{2}} \left(1 - \sqrt{2}\right)}+C$$

$$$\int x^{- \sqrt{2}}\, dx = \frac{1}{x^{-1 + \sqrt{2}} \left(1 - \sqrt{2}\right)} + C$$$A