а) f'(x) = 2x; => f(x) = x² + C
б) f'(x) = cos x; => f(x) = sin x + C
в) f'(x) = 3; => f(x) = 3x + C
г) f'(x) = -sin x; => f(x) = cos x + C
Ответ: a) f(x) = x² + C; б) f(x) = sin x + C; в) f(x) = 3x + C; г) f(x) = cos x + C