Prove that $xg(x)

Question

Let $f(x)$ satisfy the differential equation$$frac{d(f(x))}{dx}+f(x)=g(x)$$
where $f(x)$ and $g(x)$ are continuous functions. Also it is known that $f(x)$ is a decreasing function of $x$ for all positive x.
Prove that $$xg(x)<int_{0}^xg(x)dx$$
My Attempt:
Let $H(x)=xg(x)-int_{0}^xg(x)dx$
$H'(x)=xg'(x)$
What can we say about $g(x)$ and $g'(x)$.
I think some information seems to be missing. If it is given that $f(0)geq 0$ something may be worked out.

Bachamohamed · Answer

we put $$g(x)=frac{x+1}{(x-1)^{3}(x^{2}+1)}$$
We find $$xg(x)=frac{x(x+1)}{(x-1)^{3}(x^{2}+1)}$$
and $$displaystyleint_{0}^{x}frac{(x+1)}{(x-1)^{3}(x^{2}+1)}dx=frac{1}{2}(frac{1}{x-1})+frac{1}{2}(frac{1}{(x-1)^2})+frac{1}{2}tan(x)^{-1}$$
So
$$xin ({-infty},0)quad  xg(x)geqint_{0}^{x}g(x)dx$$
And from it must be the conditions of the function or the limits of integration in order to achieve that relationship.

Prove that $xg(x)<int_{0}^xg(x)dx$

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