Integral of 5cos*7x+2^x dx

1. Integrate term-by-term:

   1. The integral of an exponential function is itself divided by the natural logarithm of the base.

      $$\int 2^{x}\, dx = \frac{2^{x}}{\log{\left(2 \right)}}$$

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

      $$\int 5 \cos{\left(7 x \right)}\, dx = 5 \int \cos{\left(7 x \right)}\, dx$$

      1. Let $u = 7 x$.

         Then let $du = 7 dx$ and substitute $\frac{du}{7}$:

         $$\int \frac{\cos{\left(u \right)}}{7}\, du$$

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

            $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{7}$$

            1. The integral of cosine is sine:

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

            So, the result is: $\frac{\sin{\left(u \right)}}{7}$

         Now substitute $u$ back in:

         $$\frac{\sin{\left(7 x \right)}}{7}$$

      So, the result is: $\frac{5 \sin{\left(7 x \right)}}{7}$

   The result is: $\frac{2^{x}}{\log{\left(2 \right)}} + \frac{5 \sin{\left(7 x \right)}}{7}$

2. Now simplify:

   $$\frac{2^{x} + \frac{\log{\left(32 \right)} \sin{\left(7 x \right)}}{7}}{\log{\left(2 \right)}}$$

3. Add the constant of integration:

   $$\frac{2^{x} + \frac{\log{\left(32 \right)} \sin{\left(7 x \right)}}{7}}{\log{\left(2 \right)}}+ \mathrm{constant}$$

The answer is: