I have a function $$G=\int_0^S\mathrm{d}x\,f(x)g(x)\mathrm{e}^{-\int_0^x\mathrm{d}z\,g(z)}~.$$ If $f(x)$ was unity, the above integral could have been easily written as $$G=1-\mathrm{e}^{-\int_0^S\mathrm{d}x\,g(x)}~.$$ Now, for a general $f(x)$ reducing this into something simpler is not apparent to me. But, I have an idea on which I am stuggling: carrying out integration by parts infinite times to construct a series, which possibly reduces to something exponential. The approach: \begin{align} G=\mathrm{e}^{-\int_0^S\mathrm{d}x\,g(x)}\int_0^S\mathrm{d}x\,f(x)g(x)+\int_0^Sg(x)\mathrm{e}^{-\int_0^x\mathrm{d}z\,g(z)}\int_0^x\mathrm{d}y\,f(y)g(y)\\ =\mathrm{e}^{-\int_0^S\mathrm{d}x\,g(x)}\int_0^S\mathrm{d}x\,f(x)g(x)+\int_0^Sg(x)\mathrm{e}^{-\int_0^x\mathrm{d}z\,g(z)}\int_0^x\mathrm{d}y\,f(y)g(y)\\ =\mathrm{e}^{-\int_0^S\mathrm{d}x\,g(x)}\int_0^S\mathrm{d}x\,f(x)g(x)+\mathrm{e}^{-\int_0^S\mathrm{d}x\,g(x)}\int_0^Sg(x)\int_0^x\mathrm{d}y\,f(y)g(y)+\int_0^Sg(x)\mathrm{e}^{-\int_0^x\mathrm{d}z\,g(z)}\int_0^{x}\mathrm{d}y_1g(y_1)\int_0^{y_1}\mathrm{d}y_2\,f(y_2)g(y_2)~.\\ \end{align} At this point, it seems like there is some pattern. The final answer is desired to be in the form of $$G=\mathrm{e}^{-\int_0^S\mathrm{d}x\,g(x)}\times\text{function}\left(\int_0^S\mathrm{d}x\,f(x)g(x)\right),$$ or something similar. Any ideas on how to complete this idea?
Addition: Letting $$K(S)=\int_0^S\mathrm{d}x\,f(x)g(x),$$ The entire problem can be reduced to finding an appropriate representation for the following sum: $$K(S)+\int_0^Sg(x)K(x)\mathrm{d}x+\int_0^Sg(x)\left(\int_0^xg(x')K(x')\mathrm{d}x'\right)\mathrm{d}x+\cdots$$