How can I find $f(x)$ in below relation?
$f(x)= e^{I}$
$I = c_1 e^{-x^2/x0^2} - \int_0^\inf (\frac{c_2 (x_2 - x)}{(c_3^2+(x2-x)^2)^{\frac{5}{2}}}+ \frac{c_4(x2-x1)}{(c_3^2+(x2-x1)^2)^{\frac{7}{2}}})\: \: f(x_2) \: \: x_2 \:\:d x_2 $
in which, $c_i$ are real constants(for simplicity one can put them equal to 1).
I think, there is no chance to do this analytically. But you can get fine numerical solution if you iterate finding f[x] for each x.
NestList shows, you reach a fixed point for f[x] after enough iterations.
f[x_, x0_, x1_, fstart_, n_] := NestList[Exp[ Exp[-x^2/x0^2] - NIntegrate[((x2 - x)/(1 + (x2 - x)^2)^( 5/2) + (x2 - x1)/(1 + (x2 - x1)^2)^(7/2)) # x2, {x2, 0, \[Infinity]}, MaxRecursion -> 50]] &, fstart, n] ListLinePlot[f[1, 2, 3, 1, 100]] 
Therefore work fith FixedPoint
fp[x_, x0_, x1_, fstart_] := FixedPoint[ Exp[Exp[-x^2/x0^2] - NIntegrate[((x2 - x)/(1 + (x2 - x)^2)^( 5/2) + (x2 - x1)/(1 + (x2 - x1)^2)^(7/2)) # x2, {x2, 0, \[Infinity]}, MaxRecursion -> 50]] &, fstart, SameTest -> (Abs[#1 - #2] < 1*^-8 &)] And plot a solution
pl = Plot[fp[x, 2, 3, 1], {x, -6, 6}, Mesh -> All] 
If you generate an interpolation function
fip = Interpolation[Cases[pl, {_?NumericQ, _?NumericQ}, {6}], InterpolationOrder -> 5] you can even differentiate the found f[x]
Plot[fip'[x], {x, -6, 6}] 
I=Log[f[x]](1. equation) into the second, you get a complicated integral-equation. Unfortunately DSolve can't solve it... $\endgroup$