Integral of (x+3)e^4x dx

1. Rewrite the integrand:

   $$x e^{4} \left(x + 3\right) = x^{2} e^{4} + 3 x e^{4}$$

2. Integrate term-by-term:

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

      $$\int x^{2} e^{4}\, dx = e^{4} \int x^{2}\, dx$$

      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: $\frac{x^{3} e^{4}}{3}$

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

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

      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: $\frac{3 x^{2} e^{4}}{2}$

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

3. Now simplify:

   $$\frac{x^{2} \left(2 x + 9\right) e^{4}}{6}$$

4. Add the constant of integration:

   $$\frac{x^{2} \left(2 x + 9\right) e^{4}}{6}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \left(2 x + 9\right) e^{4}}{6}+ \mathrm{constant}$$