Finding Potential with d dimensions terms Lagrangian for a spherically-symmetric, real scalar field in
d spatial dimensions,
$$L=c_d \int r^{d-1}dr\left[ \frac{1}{2} \dot\phi^2 - \frac{1}{2}  \left(\frac{\partial \phi}{\partial r} \right)^2 -V(\phi)\right] \tag{1}$$
where $$v= m^2\phi^2$$,  $$c_d = 2π^{d/2} /Γ(d/2)$$ is the unit-
sphere volume in d dimensions.
The solution of $\phi$ is, 
$$\phi(r,t) = A(t)P(r,R)= A(t)e^{\frac{−r^2}
{R2}}\tag{2}$$
A solution for a scalar field in d dimensions 
$$\phi(r,t)= A(t) e ^\frac{-r^2}{R^2}$$
If we define a potential,
$$V = \phi^2- \phi^3+ \frac{\phi^4}{4}$$ 
then how do we get by integrating the above equation on the space d dimensions? $$V(A)= (1+\frac{d}{2R^2})A^2-\left(\frac{2}{3}\right)^\frac{d}{2} A^3+ \frac{A^4}{2^\frac{d+4}{2}}$$ 

I think I have understood what you want to achieve. You want to obtain a Lagrangian for $A(t)$ with a suitable potential. Using $\int\limits_{0}^{\infty}r^{d-1}e^{-ar^{2}} dr=\frac{1}{2} a^{-\frac{d}{2}}\Gamma\left( \frac{d}{2} \right)$ we find by inserting $\phi(r,t)=A(t)e^{-\frac{r^{2}}{R^{2}}}$ that
$$ \begin{aligned} L(A) &= c_{d} \int\limits_{0}^{\infty}  r^{d-1} \left( \frac{1}{2} \dot{A}^{2} \exp\left(-\frac{2 r^{2}}{R^{2}}\right) - \frac{1}{2}A^{2} \left(\frac{4r^{2}}{R^{4}}\right)\exp\left(-\frac{2 r^{2}}{R^{2}}\right) - A^{2} \exp\left(-\frac{2 r^{2}}{R^{2}}\right) + A^{3} \exp\left(-\frac{3 r^{2}}{R^{2}}\right) - \frac{1}{4} A^{4} \exp\left(-\frac{4 r^{2}}{R^{2}}\right) \right) dr \\ 
&= \frac{2\pi^{d/2}}{\Gamma\left( \frac{d}{2} \right)} \left( \dot{A}^{2} \frac{1}{4} \left( \frac{R^{2}}{2} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) -A^{2}  \frac{1}{R^{4}} \left( \frac{R^{2}}{2} \right)^{\frac{d+2}{2}}  \underbrace{\Gamma\left( \frac{d+2}{2} \right)}_{=\frac{d}{2}\Gamma\left( \frac{d}{2} \right)} - A^{2} \frac{1}{2} \left( \frac{R^{2}}{2} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) + A^{3} \frac{1}{2} \left( \frac{R^{2}}{3} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) -A^{4} \frac{1}{8} \left( \frac{R^{2}}{4} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) \right) \\ 
&= \pi^{d/2} R^{d} \left(  \dot{A}^{2} \frac{1}{2} \left( \frac{1}{2} \right)^{\frac{d}{2}} -A^{2}  \frac{1}{ R^{2}} \left( \frac{1}{2} \right)^{\frac{d+2}{2}} d - A^{2} \left( \frac{1}{2} \right)^{\frac{d}{2}}\ + A^{3}  \left( \frac{1}{3} \right)^{\frac{d}{2}} -A^{4} \frac{1}{4} \left( \frac{1}{4} \right)^{\frac{d}{2}} \right) \\ &= \left( \frac{\pi R^{2}}{2} \right)^{\frac{d}{2}} \left( \frac{1}{2} \dot{A}^{2}   - \left(1+ \frac{d}{2 R^{2}} \right) A^{2} + A^{3}  \left( \frac{2}{3} \right)^{\frac{d}{2}} -A^{4} \left( \frac{1}{2} \right)^{\frac{d+4}{2}} \right). \end{aligned}$$
Dropping the overall constant $\left( \frac{\pi R^{2}}{2} \right)^{\frac{d}{2}}$ one thus has the potential $V(A)=\left(1+ \frac{d}{2 R^{2}} \right) A^{2} - A^{3}  \left( \frac{2}{3} \right)^{\frac{d}{2}} +A^{4} \left( \frac{1}{2} \right)^{\frac{d+4}{2}}$ which coincides with the expression you were looking for.
Remark: You mentioned a mass term $v=m^{2}\phi^{2}$ which somehowdoes not appear in your Lagrangian. I also think you might want to add certain coupling constants to adjust the units in $V(\phi)$. However, those manipulations are easily achieved.
