I have following function of $x$ for $x>0$ $$f(x)=-(x+1)\exp(a+x)Ei(-a-x)-\frac{x}{a+x}$$ where $a>0$ and $Ei(x)$ is the exponential integral. I have plotted this function for different values of $a$ in WA and it appears that the function decreases with increasing values of $x$. Can we prove that $f(x)$ is a decreasing function of $x$? I will be very thankful to you for your help. The derivative of the above function is given as follows $$f'(x)=-\frac{(x+2)e^{a+x}(a+x)^2Ei(-a-x)+a(x+2)+x(x+1)}{(a+x)^2}$$
Edit:
Can we say anything if $0<a<1$?
Edit 2 (My goal and the way to achieve it):
Actually I am interested in finding the maximum value of $f(x)$. If could show that $f(x)$ is a decreasing function of $x$ (which I think we can show through plotting for different values of $a$ but it may not be possible to show that result analytically) then it means $f(0)$ is the maximum value. Since I can not show analytically that $f(x)$ is decreasing function therefore in the following I devise a plan to show an upper bound on the maximum achievable value of $f(x)$. First I describe the following facts.
1- $\frac{-x}{a+x}$ is a decreasing function and its range is [0,1].
2- $g(x)=-(x+1)\exp(a+x)Ei(-a-x)$ is a product of one positive increasing function (i.e. $(x+1)\exp(a+x)$) and one positive decreasing function (i.e. $-Ei(-a-x)$). Therefore $g(x)$ is a quasiconcave function and we can find the maximum value of $g(x)$. This maximum value plus $1$ will be the upper bound on the maximum achievable value of $f(x)$.
If this reasoning is wrong then please correct me. Thanks in advance.