Integral of sqrt(1/x) dx

1. Let $u = \frac{1}{x}$.

   Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$:

   $$\int \left(- \frac{1}{u^{\frac{3}{2}}}\right)\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{1}{u^{\frac{3}{2}}}\, du = - \int \frac{1}{u^{\frac{3}{2}}}\, du$$

      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \frac{1}{u^{\frac{3}{2}}}\, du = - \frac{2}{\sqrt{u}}$$

      So, the result is: $\frac{2}{\sqrt{u}}$

   Now substitute $u$ back in:

   $$\frac{2}{\sqrt{\frac{1}{x}}}$$

2. Add the constant of integration:

   $$\frac{2}{\sqrt{\frac{1}{x}}}+ \mathrm{constant}$$

The answer is:

$$\frac{2}{\sqrt{\frac{1}{x}}}+ \mathrm{constant}$$