Hello! Could you tell me about how to take the next numerical calculation in mathematica? (perhaps there are special packages). I have an expression (in reality slightly more complex):

$V=x^2 + \int_a^b l \sqrt{l^2-m^2} \left(\text Log \left(e^{-\left(\beta \left(\sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x)^2+N}+u\right)\right)}+1\right)\right) \, dl $

 V = x^2 + 
   Integrate[
     l*Sqrt[l^2 - m^2]*
     Log*(1 + 
       e^(-(\[Beta]*(u + 
         Sqrt[(Sqrt[l^2 - m^2] + U)^2 + (m + x)^2 + N])))), {l, a, b}]

where x is function of $l$; $m$, $N$ are constants; $\beta$, $u$, $U$ are parameters. I need to find the dependence $U$ on $u$ and $\beta$ (in order to draw graph) from an equation: $\frac {\partial V} {\partial x}=0$ ( $x$ will be needed to set a constant after differentiation; In reality,there is not the derivative, but a variation)

If I have an integral (without parameters) rather than equation, I would try to do the following ones:

1) to define the a region of integration (due to graphical representation of function) 2) to tabulate integrand 3) to calculate the integral that is to get a number.

Nevertheless I have the equation, which probably requires other method. I would appreciate Mathematica literature on this subject, or help.

I do not know how actual it is to calculate integral equation, but the integral can be led to another kind: $ l \sqrt{l^2-m^2} \left(\text {Log} \left(e^{-\left(\beta \left(\sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x)^2+N}+u\right)\right)}+1\right)\right) \to $ $ \left(l^2-m^2\right)^{3/2} \frac{\text{$\cosh(\beta $u)} +\exp \left(-\beta \sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x+y)^2+(q+z)^2}\right)}{\text{$\cosh (\beta $u)} -\cosh \left(\beta \sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x+y)^2+(q+z)^2}\right)} $

(x^2-m^2)^(3/2) (cosh(\[Beta]u)+exp(-\[Beta] Sqrt[(Sqrt[l^2-m^2]+U)^2+(m+x+y)^2+(z+q)^2]))/(cosh(\[Beta]u)-cosh(\[Beta] Sqrt[(Sqrt[l^2-m^2]+U)^2+(m+x+y)^2+(z+q)^2]))

Perhaps it was necessary to start with something simpler. Let

$V=x^2+\int_{a}^{b} x u U \beta l dl$

$\frac {\partial V} {\partial x}=0$

$x$ is assumed a constant after differentiating.

$\int_a^b \beta l u U \, dl+2 x=0$

$U=\frac{4 x}{\beta u \left(a^2+b^2\right)}$

In all I get the dependency $U$, on $u$ and $\beta$

I need to do the