# `f''(x) = 6x + sin(x)` Find `f`.

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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.` **