antiderivative help???

Question

[7/(x^1/3)]-[8((x^2)^1/3)]

&

[(4x^4)-2x]/(x^3)

I need the most general antiderivative without the +C part

Answer 1

Find the antiderivative of [7/(x^1/3)] - [8((x^2)^1/3)].

∫{[7/(x^1/3)] - [8((x^2)^1/3)]} dx

= ∫{[7/(x^(1/3))] - [8(x^(2/3)]} dx

= 7(3/2)x^(2/3) - 8(3/5)x^(5/3)
= (21/2)x^(2/3) - (24/5)x^(5/3)
____________________

Find the antiderivative of [(4x^4) - 2x] / (x^3).

Break it into separate fractions first, then integrate.

[(4x^4) - 2x] / (x^3) = 4x - 2/x²

Now we can integrate.

∫{[(4x^4) - 2x] / (x^3)}dx
= ∫{4x - 2/x²}dx = 2x² + 2/x

I left off the + C as requested but it still applies.

Answer 2

basic rule of integration
the antiderivate of x^n=(1/n+1)(x^(n+1))
when x is in the denominator the exponent is negative
(-1/3)+1=(-1/3)+(3/3)=2/3

7(3/2)x^(2/3)- 8(3/5)x^(5/3)
for the second function simplify it first to
4x-2(x^(-2))

the most general antiderivative means with the +c part
then take the antiderivative

Answer 3

[(21/2)x^(2/3)] - {(24/5]x^(5/3)]

&

2(x^2) + (2/x)