Lets solve the integral with Integrate and restrict to x>0,d>0. A plot shows at which d the one sided equation f[x, d] - e x^2 d
is zero in dependence of x (and e, later set to 3).
f[x_, d_] =
Integrate[z/(Exp[z] - 1), {z, x, x Sqrt[1 + d^2]},
Assumptions -> {x > 0, d > 0}]
(* -(1/2) x (-2 I (-1 + Sqrt[1 + d^2]) \[Pi] + d^2 x + 2 Log[-1 + E^x] -
2 Sqrt[1 + d^2] Log[-1 + E^(Sqrt[1 + d^2] x)]) - PolyLog[2, E^x] +
PolyLog[2, E^(Sqrt[1 + d^2] x)] *)
Manipulate[
Plot[f[x, d] - e x^2 d // Chop, {d, 0, 30},
PlotRange -> 1], {{x, .15}, 0, 1, Appearance -> "Labeled"}, {{e, 3},
0, 5, Appearance -> "Labeled"}]
In order to get the two d values where equation is zero (here called dzero1 and dzero2), regard d as a function of x, d[x], differentiate for x ( you get also d'[x]) and solve with NDSolve with two initial conditions for d (her for d[.15]) and plot result (attention: LogPlot)
dd = D[0 == f[x, d] - e x^2 d /. {e -> 3} /. d -> d[x], x] //
ExpandAll // Simplify
d01 = d /.
FindRoot[f[x, d] - e x^2 d /. {e -> 3} /. x -> .15, {d, 3}] // Chop
d02 = d /.
FindRoot[f[x, d] - e x^2 d /. {e -> 3} /. x -> .15, {d, 15}] // Chop
dzero1 = d /. First@NDSolve[{dd, d[15/100] == d01}, d, {x, 10^-6, 1}]
dzero2 = d /. First@NDSolve[{dd, d[15/100] == d02}, d, {x, 10^-6, 1}]
LogPlot[Evaluate[{dzero1[x], dzero2[x]}], {x, 10^-6, .21},
PlotRange -> 100]

NIntegrate
when there are unknowns in the integrand or in the limits as in your case. It is a numerical integration after all. You can useIntegrate
. But the resulting equation can't be solved ford
, too complicated. If you have some numerical values forx
ande
, there could be a better chance. $\endgroup$