Integral of sin(8*x)+e^(3*x) dx

1. Integrate term-by-term:

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

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

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

         $$\text{False}$$

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

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

   1. Let $u = 8 x$.

      Then let $du = 8 dx$ and substitute $\frac{du}{8}$:

      $$\int \frac{\sin{\left(u \right)}}{8}\, du$$

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

         $$\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{8}$$

         1. The integral of sine is negative cosine:

            $$\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$$

         So, the result is: $- \frac{\cos{\left(u \right)}}{8}$

      Now substitute $u$ back in:

      $$- \frac{\cos{\left(8 x \right)}}{8}$$

   The result is: $\frac{e^{3 x}}{3} - \frac{\cos{\left(8 x \right)}}{8}$

2. Add the constant of integration:

   $$\frac{e^{3 x}}{3} - \frac{\cos{\left(8 x \right)}}{8}+ \mathrm{constant}$$

The answer is:

$$\frac{e^{3 x}}{3} - \frac{\cos{\left(8 x \right)}}{8}+ \mathrm{constant}$$