Suppose I have a quotient $f(x)/g(x)$ well defined on $[-1,1]$ such that $f(0) = g(0) = 0$ and $f(0)/g(0)$. Suppose they are both differentiable functions and $f(x)/g(x)$ is continuous.
In such a situation ($1/g(x)$ by itself is not well defined on all of $[-1,1]$, even though $f$ and $f/g$ are)can one apply integration by parts? as in does this hold? $$ \int_{-1}^1 f(x) g(x)^{-1} dx = \frac{F(1)}{g(1)} - \frac{F(-1)}{g(-1)} + \int_{-1}^1 \frac{F(x)}{g(x)^2}g'(x) dx $$ here $F(x) = \int_{-1}^x f(t) dt$. Thank you!
(Alternatively, my question is under what conditions can one apply integration by parts to quotients of functions)
Edit. Error has been corrected based on a solution below
The differentiation rule for quotients is $$\left(\frac{F(x)}{g(x)}\right)' = \frac{F'(x)g(x)-F(x)g'(x)}{g^2(x)}.$$
By rearranging the terms (we solve for the first term on the right hand side), this becomes $$\frac{F'(x)}{g(x)} = \left(\frac{F(x)}{g(x)}\right)' + \frac{F(x)g'(x)}{g^2(x)}.$$
Integration of this equation from $-1$ to $1$ yields $$\int_{-1}^1 \frac{F'(x)}{g(x)}dx = \frac{F(1)}{g(1)} - \frac{F(-1)}{g(-1)} + \int_{-1}^1 \frac{F(x)g'(x)}{g^2(x)}dx.$$
This is analogous to integration by parts for products. We could have also derived this equation from the usual integration by parts formula by noting that $\left(\frac{1}{g(x)}\right)'=-\frac{g'(x)}{g^2(x)}$.
Now we can set $F(x)=\int_{-1}^xf(t)dt$ and obtain the correct version of your formula: $$\int_{-1}^1 \frac{f(x)}{g(x)}dx = \frac{F(1)}{g(1)} - \frac{F(-1)}{g(-1)} + \int_{-1}^1 \frac{F(x)g'(x)}{g^2(x)}dx.$$