Integral of x^3/(x+1) dx

1. Rewrite the integrand:

   $$\frac{x^{3}}{x + 1} = x^{2} - x + 1 - \frac{1}{x + 1}$$

2. Integrate term-by-term:

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

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

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

      $$\int \left(- x\right)\, dx = - \int x\, dx$$

      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: $- \frac{x^{2}}{2}$

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

      $$\int 1\, dx = x$$

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

      $$\int \left(- \frac{1}{x + 1}\right)\, dx = - \int \frac{1}{x + 1}\, dx$$

      1. Let $u = x + 1$.

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

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(x + 1 \right)}$$

      So, the result is: $- \log{\left(x + 1 \right)}$

   The result is: $\frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log{\left(x + 1 \right)}$

3. Add the constant of integration:

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

The answer is:

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