whats the anti derivative of 1/(x^3)?? i have an answer but shows its wrong//// help?

Question

anonymous · Answer

my answer is x^-2/2

anonymous · Answer

When doing integrals with a power on the bottom, its often helpful to rewrite it like so

anti derivative of x^-3

So lets use the general rule for integration, -3 + 1 = -2

So the anti derivative would be x^(-2) / -2 or in proper form

1 / 2*x^2

anonymous · Answer

I'm sorry, I mistyped that. 

The answer is

-2 / x^2

amistre64 · Answer

x^-3 -> x^-4/4:

1/(4x^4)

anonymous · Answer

Wait I keep messing that up....

1 / -2*x^2

amistre64 · Answer

hah!!... i missed it too lol

amistre64 · Answer

-1/2x^2  is right ;)

anonymous · Answer

thanks you haha ... so how did you get that>?

amistre64 · Answer

its the reverse of the power rule of a derivative.

anonymous · Answer

got it .. thanks

amistre64 · Answer

$$D(X^n) = n * X^{n-1} \rightarrow D(x^5) = 5x^4$$

dumbcow · Answer

you do it correctly you just forgot to include the negative when you divided by the exponent

amistre64 · Answer

$$\int\limits_{} X^n dx \rightarrow \frac{X^{n+1}}{n+1}$$

anonymous · Answer

ahh i got it ,.,.,.

anonymous · Answer

thanks