Roman C. answered 10/31/15
Tutor
5.0
(851)
Masters of Education Graduate with Mathematics Expertise
∫ [-9h′(6x+y) + 9g′(6x−y)] dx
= (-3/2) ∫ [h′(6x+y)·6 + g′(6x−y)·6] dx
= (-3/2) [h′(6x+y)·(6x+y)′ + g′(6x−y)·(6x-y)′] dx
= (-3/2)[h(6x+y) + g(6x−y)] + f(y)
Here f(y) plays the role of the constant of integration. It can be any function of y since it is independent of x.