@Carl Love 

thank you very much for replying but I can not do it :( in loop can you explain it by example or correct indexed subscripted names in my program 

 thank you for helping

@Markiyan Hirnyk 

oh 

I'm so sorry I've downloaded the wrong file I have corrected  it now.

@vv 

thank you very much but where use simplify in what step? and I use the mathematical method adomian decomposition it should the function Growing naturally when to use the other method to solve the integral equation 

@vv 

but not true values :( 

@Christopher2222  

thank you very much for helping but the final values are 

alpha=4 and beta=0

I can not find it :(

@Markiyan Hirnyk 

thank you very much but I have a question How to draw the figure inside the pox?

@vv 

Thank you very much for your helping but the equation of Abel integral also I can not find the solution 

> f := proc (x) options operator, arrow; (4/3)*sin(x)^(3/4) end proc; g := proc (x) options operator, arrow; sin(x) end proc; alpha := 1/4; eval(diff(g(x), x), x = t); k := proc (x, t) options operator, arrow; 1/(g(x)-g(t))^alpha end proc;

> j := int((eval(diff(g(x), x), x = t))*f(t)/(g(x)-g(t))^(1-alpha), t = 0 .. x); int((eval(diff(g(x), x), x = t))*f(t)/(g(x)-g(t))^(1-alpha), t = 0 .. x); `assuming`([J = limit(%, t = x, left)-(limit(%, t = 0, right))], [0 < x and x < (1/2)*Pi, cos(x) <> 0]);

> f := proc (x) options operator, arrow; (4/3)*sin(x)^(3/4) end proc; g := proc (x) options operator, arrow; sin(x) end proc; alpha := 1/4; eval(diff(g(x), x), x = t); k := proc (x, t) options operator, arrow; 1/(g(x)-g(t))^alpha end proc;                             cos(t)> j := int((eval(diff(g(x), x), x = t))*f(t)/(g(x)-g(t))^(1-alpha), t = 0 .. x); int((eval(diff(g(x), x), x = t))*f(t)/(g(x)-g(t))^(1-alpha), t = 0 .. x); `assuming`([J = limit(%, t = x, left)-(limit(%, t = 0, right))], [0 < x and x < (1/2)*Pi, cos(x) <> 0]);

@Preben Alsholm 
thanks alot of  your advice  but how change the limit of integral ?

@Carl Love 

tanks alot for advice but I want integrate both side of integro - diff from 0 to x to convert it to volterra integral equation then solve by the mehod of volterra integral equation 

I can not integrate both side from 0 to x and subs in initial condition

@Preben Alsholm

thank you very much to your advice Iam  trying and successful

Thanks for the reply i am try this and successful :)

Thank you to answer, but a program that takes a long time at the expense of integration because it is complicated Is there a way to reduce the size of calculations using ADM for non linear quadraic integral equation 

> k := proc (s, t) options operator, arrow; (1/2)*t^2/exp(t+s) end proc;
f := proc (t) options operator, arrow; exp(-t) end proc;
X[0] := f(t);
> F(x):=ln(1+s*x):
N:=10:
> for n from 0 to 5 do AD[n] := (diff(F(sum(lambda^i*s[i], i = 0 .. 3)), [`$`(lambda, n)]))/factorial(n);
A[n] := subs(lambda = 0, AD[n]) end do;

for i to 5 do X[i] := X[i-1]*t^2*(int(A[i-1]/exp(s), s = 0 .. t))/(2*exp(t));
s[i] := subs(t = s, X[i]) end do