Integral of (4x+3)^3 dX

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 4 x + 3$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{u^{3}}{16}\, du$$

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

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

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

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

         So, the result is: $\frac{u^{4}}{16}$

      Now substitute $u$ back in:

      $$\frac{\left(4 x + 3\right)^{4}}{16}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(4 x + 3\right)^{3} = 64 x^{3} + 144 x^{2} + 108 x + 27$$

   2. Integrate term-by-term:

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

         $$\int 64 x^{3}\, dx = 64 \int x^{3}\, dx$$

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

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

         So, the result is: $16 x^{4}$

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

         $$\int 144 x^{2}\, dx = 144 \int x^{2}\, dx$$

         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: $48 x^{3}$

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

         $$\int 108 x\, dx = 108 \int x\, dx$$

         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: $54 x^{2}$

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

         $$\int 27\, dx = 27 x$$

      The result is: $16 x^{4} + 48 x^{3} + 54 x^{2} + 27 x$

2. Now simplify:

   $$\frac{\left(4 x + 3\right)^{4}}{16}$$

3. Add the constant of integration:

   $$\frac{\left(4 x + 3\right)^{4}}{16}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(4 x + 3\right)^{4}}{16}+ \mathrm{constant}$$