Integral of sqrt(x-5) dx

1. Let $u = x - 5$.

   Then let $du = dx$ and substitute $du$:

   $$\int \sqrt{u}\, du$$

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

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

   Now substitute $u$ back in:

   $$\frac{2 \left(x - 5\right)^{\frac{3}{2}}}{3}$$

2. Now simplify:

   $$\frac{2 \left(x - 5\right)^{\frac{3}{2}}}{3}$$

3. Add the constant of integration:

   $$\frac{2 \left(x - 5\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{2 \left(x - 5\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$