Integration type problem Let $f(x)$ and $g(x$) be continuous, positive function such that $f(–x) = g(x) – 1$,$f(x)=\frac{g(x)}{g(-x)}$ and $\int\limits_{ - 20}^{20} {f\left( x \right)dx}  = 2020$
then the value of is $\int\limits_{ - 20}^{20} {\frac{{f\left( x \right)}}{{g\left( x \right)}}dx} $ is
(A) 1010
(B) 1050
(C) 2020
(D) 2050
My approach is as follow
GIven $f\left( { - x} \right) = g\left( x \right) - 1\& f\left( x \right) = \frac{{g\left( x \right)}}{{g\left( { - x} \right)}}$
$\frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{1}{{g\left( { - x} \right)}}\& f\left( { - x} \right) = g\left( x \right) - 1 \Rightarrow f\left( x \right) + 1 = g\left( { - x} \right) \Rightarrow \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{1}{{f\left( x \right) + 1}}$
$\int\limits_{ - 20}^{20} {\frac{{f\left( x \right)}}{{g\left( x \right)}}dx}  \Rightarrow \int\limits_{ - 20}^{20} {\frac{1}{{f\left( x \right) + 1}}dx} $
Not able to approach from here

So firstly we make the computation
$$\frac{f(x)}{g(x)}=\frac{1}{g(-x)}=\frac{1}{1+f(x)}.$$
Now notice that
$$\int_{-20}^{20}\frac{f(x)}{g(x)}~\mathrm{d}x=\int_{-20}^{20} \frac{\mathrm{d}x}{1+f(x)}=\int_{-20}^{0} \frac{\mathrm{d}x}{1+f(x)}+\int_{0}^{20} \frac{\mathrm{d}x}{1+f(x)}=\int_{0}^{20} \frac{\mathrm{d}x}{1+f(-x)}+\int_{0}^{20} \frac{\mathrm{d}x}{1+f(x)}=\int_0^{20}\frac{2+f(x)+f(-x)}{1+f(x)+f(-x)+f(x)f(-x)}~\mathrm{d}x.$$
However using the fact that $f(x)=\frac{g(x)}{g(-x)}$, we can easily see that
$$f(x)f(-x)=1,$$
and so
$$\int_{-20}^{20} \frac{f(x)}{g(x)}~\mathrm{d}x=\int_0^{20}\frac{2+f(x)+f(-x)}{2+f(x)+f(-x)}~\mathrm{d}x=\int_0^{20}\mathrm{d}x=20.$$
So either there is a mistake in my computation, or none of the answers are correct.
