Integral of (x+1)^3 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x + 1$.

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

      $$\int u^{3}\, du$$

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

         $$\int u^{3}\, du = \frac{u^{4}}{4}$$

      Now substitute $u$ back in:

      $$\frac{\left(x + 1\right)^{4}}{4}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(x + 1\right)^{3} = x^{3} + 3 x^{2} + 3 x + 1$$

   2. Integrate term-by-term:

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

         $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

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

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

         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: $x^{3}$

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

         $$\int 3 x\, dx = 3 \int x\, dx$$

         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: $\frac{3 x^{2}}{2}$

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

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

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

2. Now simplify:

   $$\frac{\left(x + 1\right)^{4}}{4}$$

3. Add the constant of integration:

   $$\frac{\left(x + 1\right)^{4}}{4}+ \mathrm{constant}$$

The answer is: