Code:
Plot[Integrate[1/Sqrt[(x - 3.2445)^(1/2)]
\!\(\*SuperscriptBox[\(\[ExponentialE]\), \(\[Integral]\((x -
3.2446)\)^\((1/4)\) \[DifferentialD]x\)]\), {x, -10, 10}]]
What do you really want to do??? That is by no means clear: If you do an integral
Integrate[ … , { x, a, b } ]
you get a single number - if you are lucky. Here this would not work, because your integrand has a singularity for x == 3.2445 !
And in addition: how could one plot a single number? You are supposed to have a function of something.
I try to guess what you perhaps mean.
I define a function which resemble yours (but I am not sure if you meant this)
f[x_, low_] :=
NIntegrate[NIntegrate[Exp[(z - 3.2446)^(1/4)], {z, low, x}]/Sqrt[(y - 3.2445)^(1/2)], {y, low, x}]
Left of the singularity you get complex numbers
In[37]:= f[3, -10]
Out[37]= 321.353 + 115.379 I
But in the positive region and above the singularity you can get a plot:
Plot[f[x, 3.3], {x, 3.3, 10}]
Somehow my message has dissappeared. I have no time yet to type it again. So in short
f[x_, low_] :=
NIntegrate[NIntegrate[Exp[(z - 3.2446)^(1/4)], {z, low, y}]/ Sqrt[(y - 3.2445)^(1/2)], {y, low, x}]
p1 = Plot[f[x, 3.3], {x, 3.3, 10}]
Perhaps you intended to plot the result of the indefinite integral.
You might have fewer difficulties with the evaluation of the integration interacting with the evaluation of the plot if you do this in two steps
fun=Integrate[E^Integrate[(x-3.2446)^(1/4),x]/(x-3.2446)^(1/4),x]
and then as a separate step
Plot[{Re[fun],Im[fun]}, {x,-10,10}]
I plotted the real and imaginary parts on the same graph because for some values of x your fourth roots resulted in complex values