It might be a easy question for you :)
Set the two variables as $u $ and $v$ $s.t.$ $u=g(x)$ and $v=h(x)$
Let the differentiable function $f(x,t)$ and its antiderivative $F(x)(=F(u,v,x))=\int_{u}^{v}f(x,t) dt$
what is $F_x(x)$ ????
($F_x$ and $f_x$ mean partial derivative respectively)
There is some abuse of notation in the setup of your question. A priori we have the function $$\Phi:\>{\mathbb R}^3\to{\mathbb R},\qquad (u,v,w)\mapsto\Phi(u,v,w):=\int_u^v f(w,t)\>dt$$ of the three variables $u$, $v$, and $w$. One computes $${\partial\Phi\over\partial u}=-f(w,u),\quad{\partial\Phi\over\partial v}=f(w,v),\quad{\partial\Phi\over\partial w}=\int_u^v f_{.1}(w,t)\>dt\ ,$$ where $f_{.1}$ denotes the partial derivative of $f$ with respect to its first variable.
In addition we have a function $$\psi:\quad x\mapsto \psi(x):=\bigl(g(x),h(x),x\bigr)\in{\mathbb R}^3$$ with $\psi'(x)=\bigl(g'(x),h'(x),1\bigr)$.
Compose $\psi$ and $\Phi$ to the function $$x\mapsto F(x):=\Phi(\psi(x)\bigr)=\int_{g(x)}^{h(x)} f(x,t)\>dt\ .$$ By the chain rule we get $$\eqalign{F'(x)&=d\Phi\bigl(\psi(x)\bigr).\psi'(x)\cr &={\partial\Phi\over\partial u}\biggr|_{(g(x),h(x),x)}\>g'(x)+{\partial\Phi\over\partial v}\biggr|_{(g(x),h(x),x)}\>h'(x)+{\partial\Phi\over\partial w}\biggr|_{(g(x),h(x),x)}\>1\cr &=-f\bigl(x,g(x)\bigr)g'(x)+f\bigl(x,h(x)\bigr)h'(x)+\int_{g(x)}^{h(x)}f_{.1}(x,t)\>dt\ ,\cr}$$