# Thread: Aren't derivative and integral inverse operations?

1. ## Aren't derivative and integral inverse operations?

Hey what's going on here. I thought if you take the derivative of a function, then integrate that derivative, you get the original function back.

But when I perform this routine on the function below I get a different function from the original. I'd appreciate it if someone could point out where I've gone astray. Thanks.

$\displaystyle f(x)=\frac{x^2}{x^2+1}$
$\displaystyle \frac{df}{dx}=\frac{2x}{(x^2+1)^2}$
$\displaystyle \int\frac{df}{dx}dx=\frac{-1}{x^2+1}$

2. They are inverses - up to a constant. Your f(x) and the result of your integration differ by a constant (1, in this case). That is, if you add 1 to the result of your integration, you will get the function with which you started.