Integral of 8*x^2+16*x+17*cos4x dx

1. Integrate term-by-term:

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

      $$\int 8 x^{2}\, dx = 8 \int x^{2}\, dx$$

      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: $\frac{8 x^{3}}{3}$

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

      $$\int 16 x\, dx = 16 \int x\, dx$$

      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: $8 x^{2}$

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

      $$\int 17 \cos{\left(4 x \right)}\, dx = 17 \int \cos{\left(4 x \right)}\, dx$$

      1. Let $u = 4 x$.

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

         $$\int \frac{\cos{\left(u \right)}}{16}\, du$$

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

            $$\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$$

            1. The integral of cosine is sine:

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

            So, the result is: $\frac{\sin{\left(u \right)}}{4}$

         Now substitute $u$ back in:

         $$\frac{\sin{\left(4 x \right)}}{4}$$

      So, the result is: $\frac{17 \sin{\left(4 x \right)}}{4}$

   The result is: $\frac{8 x^{3}}{3} + 8 x^{2} + \frac{17 \sin{\left(4 x \right)}}{4}$

2. Add the constant of integration:

   $$\frac{8 x^{3}}{3} + 8 x^{2} + \frac{17 \sin{\left(4 x \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{8 x^{3}}{3} + 8 x^{2} + \frac{17 \sin{\left(4 x \right)}}{4}+ \mathrm{constant}$$