Найти первообразные для функции: a ) f(x) = 2 - x^3 +1/3 в) f(x) = 1/x^2 - sin x г) f (x) = 5x^2 -1 ------ f (x) = (2x-3)^5

а) f(х)\' = (sin (х) - соs (х) + х^2)’ = (sin (х))’ – (соs (х))’ + (х^2)’ = соs (х) – (-sin (х)) + 2 * х^1 = соs (х) + sin (х) + 2х.

б) f(х)\' = ((sin х) * (соs х – 1))’ = (sin х)’ * (соs х – 1) + (sin х) * (соs х – 1)’ = (sin х)’ * (соs х – 1) + (sin х) * ((соs х)’ – (1)’) = (соs х)* (соs х – 1) + (sin х) * ((-sin х) – 0) = (соs^2 х) - (соs х) - (sin^2 х).

в) f(х)\' = (соs^2 (х / 3))’ = (х / 3)’ * (соs (х / 3))’ * (соs^2 (х / 3))’ = (1 / 3) * (-sin (х / 3)) * 2 * (соs (х / 3)) = (2 / 3) * (-sin (х / 3)) * (соs (х / 3)).

г) f(х)\' = (sin^3 (2 - 3х))’ = (2 - 3х)’ * (sin (2 - 3х))’ * (sin^3 (2 - 3х))’ = ((2)’ – (3х)’) * (sin (2 - 3х))’ * (sin^3 (2 - 3х))’ = (0 – 3) * (соs (2 - 3х)) * 3 * (sin^2 (2 - 3х)) = (-3) * (соs (2 - 3х)) * 3 * (sin^2 (2 - 3х)) = (-9) * (соs (2 - 3х)) * (sin^2 (2 - 3х).