Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 28

You need to evaluate f, knowing the second derivative, hence, you need to use the following principle, such that:
int f''(x)dx = f'(x) + c
int f'(x) dx = f(x) + c
Hence, you need to evaluate the indefinite integral of second derivative, such that:
int (6x + sin x)dx = int 6x dx + int sin x dx
int (6x + sin x)dx = 6x^2/2 - cos x + c
int (6x + sin x)dx = 3x^2 - cos x + c
Hence, one of the first derivatives is f'(x) = 3x^2 - cos x. + c
You need to evaluate the function, such that:
int (3x^2 - cos x + c)dx = int 3x^2 dx - int cos x dx + int c dx
int (3x^2 - cos x + c)dx =3x^3/3 - sin x + cx + c
int (3x^2 - cos x)dx =x^3 - sin x + cx + c
Hence, evaluating the function, yields f(x) = x^3 - sin x + cx + c.