How to integrate $\int_{0}^{1}g'(x)f(g(x),x)dx$

Question

I know how to integrate $$\int_{0}^{1}g'(x)f(g(x))dx$$ for $g$ differentiable, I just need to take $s = g(x)$ and then I find $$\int_{g(0)}^{g(1)}f(s)ds$$ But what happens when $f$ depends explicitely on $x$, i.e. how to do the integration for $$ \int_{0}^{1} g'(x)f(g(x),x)dx$$
For now, my only idea is to write this integral as $$ \int_{0}^{1}(g'(x), 1). (f(g(x),x),0)dx $$ so that I have $(g(x),x)'$. But it does not seem to work ...

Answer 1

After some work, I think I found the answer. So, I have $$ \int_{0}^{1}g'(x)f(g(x),x)dx = \int_{0}^{1}(g'(x),1).(f(g(x),x),0)dx$$ I define the function $L : \mathbb{R}^{2}\to \mathbb{R}^{2}$ by $$ L(\gamma (x)) = (f(\gamma (x)), 0)$$ where $\gamma (x) = (g(x),x)$. The problem is therefore to integrate $$ \int_{0}^{1}\gamma '(x) L(\gamma (x))dx $$ At this point, it looks very similar to $\int g' f(g)$. I set $s = (s_{1},s_{2}) = \gamma (x)$, which gives us $ds = (ds_{1},ds_{2}) = \gamma'(x) dx$. The previous integral becomes $$ \int_{\gamma (0)}^{\gamma (1)} L(s)ds = \int_{\gamma (0)}^{\gamma (1)}(f(s),0)ds$$ which concludes the proof.