inverse of function defined by an integral Sorry for any inconveniences with this question, I am new here and I am an engineer, not a mathematician (booo! :D).
I have an equation like that
$$g(x)=\int_x^1f(r)\cdot \frac{1}{\sqrt{r^2+x^2}}dr$$
$g(x)$ is known on the relevant interval for $x\in [0,1]$, smooth, strictly decreasing and positive. I want to know the function $f$, but I have no clue where to start
I am thankful for any help
 A: we have:
$$g(x)=\int_x^1f(r)\frac{1}{\sqrt{r^2+x^2}}dr$$
now if we differentiate wrt. $x$ we get:
$$g'(x)=-\frac{f(x)}{\sqrt{2}x}-\int_x^1f(r)\frac{x}{(r^2+x^2)^{3/2}}dr$$
I will now try and address this integral:
$$I=\int_x^1f(r)\frac{x}{(r^2+x^2)^{3/2}}dr=-\left[\frac{f(r)}{\sqrt{r^2+x^2}}\right]_{r=x}^1+\int_x^1f'(r)\frac{dr}{\sqrt{r^2+x^2}}$$
if we combine this we get:
$$g'(x)=-\frac{f(x)}{\sqrt{2}x}+\frac{f(1)}{\sqrt{x^2+1}}-\frac{f(x)}{\sqrt{2}x}-\int_x^1f'(r)\frac{dr}{\sqrt{r^2+x^2}}$$
this isn't ideal but this gives $f(x)$ in terms of $g'(x)$ and an integral of $f'(x)$

Alternatively we can use bounds, notice that:
$$\sqrt{2}x\le\sqrt{r^2+x^2}\le\sqrt{x^2+1}$$
$$\frac{1}{\sqrt{x^2+1}}\le\frac{1}{\sqrt{r^2+x^2}}\le \frac{1}{\sqrt{2}x}$$
and so:
$$\frac{1}{\sqrt{x^2+1}}\int_x^1 f(r)dr\le g(x)\le \frac{1}{\sqrt{2}x}\int_x^1f(r)dr$$
i.e.;
$$(F(1)-F(x))\frac{1}{\sqrt{x^2+1}}\le g(x)\le (F(1)-F(x))\frac{1}{\sqrt{2}x}$$
if we now split the inequality up we get:
$$F(1)-F(x)\le\sqrt{x^2+1}g(x)\Rightarrow\,-f(x)\le \frac{x}{\sqrt{x^2+1}}g(x)+\sqrt{x^2+1}g'(x)$$
and do the same for the other side
