Bonus: Find the integral of e^2f(x)

If we let u = f(x) then du = f '(x)dx. So the integral becomes the itegral of u²du which becomes u³/3 or f³(x)/3 + C. In the general case, the integral solution becomes fⁿ⁺¹/(n+1) + C.
For the bonus, we have to use integration by parts (which we could have also used for the problem itself). We know the integral of u dv = u*v - integ(v du). So we set up u = f(x), dv = e^2f(x) f'(x) dx. Thus, du = f'(x) dx, and v can be found using the same technique we did for the previous part of our problem, so v becomes e^2f(x)/2.
Hence, uv - integ(v du) yields to f(x)*e^2f(x)/2 - integ(e^2f(x)/2 f'(x) dx. This becomes f(x)*e^2f(x)/2 - e^2f(x)/4 or more simply e^2f(x) * [f(x)/2 - 1/4].