Find the most general antiderivative for the function....?

Question

(7)/(cubicroot(x))-3*cubicroot(x^2)

I know that you can rewrite the cubic roots to x^(1/3) but I am confused on solving it. I also know about the + C on the end of the final answer as well.

I know it dissappers. But, if you put you mouse over it without clicking on it you will see the rest of the question.

Answer 1

Rewrite it as 7x^(-1/3) - 3x^(2/3), then integrate:
∫7x^(-1/3) - 3x^(2/3) dx

= 7 x^(2/3) / (2/3) - 3x^(5/3) / (5/3) + c

= 21/2 x^(2/3) - (9/5)x^(5/3) + c

Answer 2

write it as
7*(-cubicroot(x))-3*(cubicroot(x)) and find the antiderivative.
for the negative cubicroot, you will have to add 1 to the power -1/3 which is positive 2/3. now figure out what times 2/3 equals 7
2/3x=7
10.5=x
10.5(x^(2/3) for the first part.
now add 1 to 1/3 for the positive cubicroot
now what times 4/3 = -3
4/3x = -3
x= -2.25 so the whole thing is
10.5(x^(2/3))-2.25(x^(4/3)) + C

Answer 3

f(x) = (7)/(cubicroot(x))-3*cubicroot(x^2)
f(x) = 7/(x^1/3) - 3*x^2/3

F(x) =∫7/(x^1/3) - 3*x^2/3 dx
F(x) =∫7/(x^1/3)dx -∫3*x^2/3 dx
F(x) =7∫(x^-1/3)dx - 3∫x^2/3 dx
F(x) =7(3/2)(x^2/3) - 3(3/5)x^5/3 + c
F(x) =(21/2)(x^2/3) - (9/5)x^5/3 + c

F(x) =10.5(x^2/3) - 1.8*(x^5/3) + c

₢

Answer 4

You need to edit your question ... it went "dot dot dot" and the question turned invisible.