# Solve an algebraic equation with an integral

I am trying to compute for the variable `zm` in terms of `t` which is written as an algebraic equation with an integral in it. The final answer should be `zm = zm[t]`.

`t - Integrate[(c[zm] z^(d - 1))/(f[z] Sqrt[f[z] + c[zm]^2 z^(2 d - 2)]), {z, 0, zm}] == 0`

Just a note, `c[zm]` contains a negative sign inside the square root so that `zm` must be greater than `zh` in order for `c[zm]` to be real.

``d = 3; zh = 2; c[zm_] := Sqrt[-(1 - zm^(d + 1)/zh^(d + 1))]/zm^(d - 1); f[z_] := (1 - z^(d + 1)/zh^(d + 1));  In[8]:= Integrate[(c[zm] z^(d - 1))/(f[z] Sqrt[f[z] + c[zm]^2 z^(2 d - 2)]), {z, 0, zm},   Assumptions -> zm > 2]  Out[8]= (1/64)*Pi*((-32 - 32*I) - (Sqrt[2*Pi]*zm*Sqrt[-16 + zm^4]*(-1 + Hypergeometric2F1[-(1/4), 1, 1/4, 16/zm^4]))/Gamma[5/4]^2)  Solve[t - Integrate[(c[zm] z^(d - 1))/(f[z] Sqrt[f[z] + c[zm]^2 z^(2 d - 2)]), {z, 0, zm}] == 0 , zm] ``

I am not sure if using `Solve` can really find the expression, also the result of the integral contains an imaginary term, but it should not right since c[zm] is real from the `Assumptions -> zm>2`?