Integral of xdx/√4-x^2 dx

1. Integrate term-by-term:

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

      $$\int \left(- x^{2}\right)\, dx = - \int x^{2}\, dx$$

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

         $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

      So, the result is: $- \frac{x^{3}}{3}$

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

      $$\int x 1 \cdot \frac{1}{\sqrt{4}}\, dx = \frac{\int x\, dx}{2}$$

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

         $$\int x\, dx = \frac{x^{2}}{2}$$

      So, the result is: $\frac{x^{2}}{4}$

   The result is: $- \frac{x^{3}}{3} + \frac{x^{2}}{4}$

2. Now simplify:

   $$\frac{x^{2} \cdot \left(3 - 4 x\right)}{12}$$

3. Add the constant of integration:

   $$\frac{x^{2} \cdot \left(3 - 4 x\right)}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \cdot \left(3 - 4 x\right)}{12}+ \mathrm{constant}$$