# Thread: Most general antiderivative of a function

1. ## Most general antiderivative of a function

g(x) = (5-4x^3 + 2x^6) / x^6

g'(x) = (5 - 4x^3 + 2x^6) (x^-6) + C

g'(x) = (5-(4x^4)/4 + (2x^7)/7) (x^-5)/-5 + C

g'(x) = (5 - x^4 + 2/7x^7) / (5x^5) + C

g'(x) = (-x^4 + 2/7x^7) / (x^5) + C

2. There is at least one mistake per line, both algebra and calculus errors. Try this:

$\displaystyle g(x)=(5-4x^{3} + 2x^{6}) / x^{6}=5x^{-6}-4x^{-3}+2.$

So far, I haven't done any calculus whatsoever, and thus there should be no constant of integration yet. In addition, the title of your post indicated that you are taking an antiderivative. It puzzled me, then, when I saw the notation g'(x), which is the notation for a derivative.

In any case, you can now integrate term-by-term. What do you get?

3. Thanks for your reply Ackbeet. You are right, I am trying to integrate g(x) = (5-4x^3 + 2x^6) / x^6 and should not have used g prime.

g(x) = (5-4x^3 + 2x^6) / x^6

= (5/x^6) - [(4x^3)/x^6] + [(2x^6)/x^6] + C

= (5x^-6) - (4x^3)(x^-6) + (2x^6)(x^-6) + C

= (5x^-6) - (4x^-3) + 2 + C

= [(5x^-5)/-5] - [(4x^-2)/-2] + 2x + C

= (-x^-5) + (2x^-2) + 2x + C

= (-1/x^5) + (2/x^2) + 2x + C

Is this correct?

4. Originally Posted by sparky
Thanks for your reply Ackbeet. You are right, I am trying to integrate g(x) = (5-4x^3 + 2x^6) / x^6 and should not have used g prime.

g(x) = (5-4x^3 + 2x^6) / x^6

= (5/x^6) - [(4x^3)/x^6] + [(2x^6)/x^6] + C
NO C! you don't have an added constant of integration until after you integrate! Ackbeet told you that.

= (5x^-6) - (4x^3)(x^-6) + (2x^6)(x^-6) + C

= (5x^-6) - (4x^-3) + 2 + C

= [(5x^-5)/-5] - [(4x^-2)/-2] + 2x + C
This is the first line on which "+ C" should appear. But you shouldn't have "= " in front of it. That means it is equal to the same as all the previous lines, g(x), and it isn't.
You should have
$\displaystyle \int g(x) dx= -x^{-5}-+ 2x^{-2}+ 2x+ C$

= (-x^-5) + (2x^-2) + 2x + C

= (-1/x^5) + (2/x^2) + 2x + C

Is this correct?

5. Thanks alot HallsofIvy!