Integral of x/sqrt(1+x) dx

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

   Then let $du = \frac{dx}{2 \sqrt{x + 1}}$ and substitute $du$:

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

   1. Integrate term-by-term:

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

         $$\int 2 u^{2}\, du = 2 \int u^{2}\, du$$

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

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

         So, the result is: $\frac{2 u^{3}}{3}$

      1. The integral of a constant is the constant times the variable of integration:

         $$\int \left(-2\right)\, du = - 2 u$$

      The result is: $\frac{2 u^{3}}{3} - 2 u$

   Now substitute $u$ back in:

   $$\frac{2 \left(x + 1\right)^{\frac{3}{2}}}{3} - 2 \sqrt{x + 1}$$

2. Now simplify:

   $$\frac{2 \left(x - 2\right) \sqrt{x + 1}}{3}$$

3. Add the constant of integration:

   $$\frac{2 \left(x - 2\right) \sqrt{x + 1}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{2 \left(x - 2\right) \sqrt{x + 1}}{3}+ \mathrm{constant}$$