Integral of sqrt(x)*i*n*x dx

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

   Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $2 i du n$:

   $$\int 2 i n u^{4}\, du$$

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

      $$\int u^{4}\, du = 2 i n \int u^{4}\, du$$

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

         $$\int u^{4}\, du = \frac{u^{5}}{5}$$

      So, the result is: $\frac{2 u^{5} i n}{5}$

   Now substitute $u$ back in:

   $$\frac{2 i n x^{\frac{5}{2}}}{5}$$

2. Add the constant of integration:

   $$\frac{2 i n x^{\frac{5}{2}}}{5}+ \mathrm{constant}$$

The answer is:

$$\frac{2 i n x^{\frac{5}{2}}}{5}+ \mathrm{constant}$$