ugh cant get this one? ...Evaluate the indefinite integral?

Question

5sqrt(2-5y) dy

Thats the fifth root of 2-5y.

I got 1-(x^4)^(9)

but its wrong.

Answer 1

INT[(2-5y)^(1/5)dy]

Let u = 2-5y, then du = -5dy, so dy = (-1/5)du

The problem becomes:

INT[u^(1/5)*(-1/5)]du = -1/5*INT(u^(1/5))du

= (-1/5)*(5/6)*u^(6/5) + C

= (-1/6) (2-5y)^(6/5) + C

Answer 2

i dont think those are the answers.

easy way to check
take the derivative of the answer,
which according to the other people is -(1/6)(2 - 5y)^(6/5) + c,
is 5(2-5y)^(1/5), which isnt what he was asking. taking the fifth root into account, the answer is (-1/3) (2-5y)^(3/2) + C

Answer 3

[(2- 5y)^(1/5) dy

let 2 - 5y = u
-5dy = du
dy = -(1/5) du
substituting
[(u)^(1/5) (-1/5) du
-(1/5)[u^(1/5) du
-(1/5)(u^(1/5 + 1)/(1/5 + 1))
-(1/5)(u^(6/5)/6/5)
-(1/6)(u^(6/5)) + c
-(1/6)(2 - 5y)^(6/5) + c

Answer 4

(-1/6)(2-5y)((2-5y)^(1/5))+C. I used "u substitution" to solve.