Calculating electrostatic potential A continuous charge distribution is spherically symmetric and has a volume charge density
$$\rho(r) = \rho_oe^{−\alpha r}$$ 
I need to find the potential as a function of '$r$' i.e. $V(r)$.
It seems fairly straight forward. I can easily find the Electric field as a function of $r$ using Gauss's Law, and then integrate the negative of the field from infinity to $r$ to get the potential. But it turns out that calculating the integrals is a tough task. Is there a simpler way to solve this problem?
 A: No, that is the simplest way to solve the problem. As mentioned in the comments, this is in the absolute scale of things a very easy problem: the spherical symmetry allows you to even have an integral to calculate, and the exponential is not only exactly integrable, but easily so. If you allow general spherically symmetric charge densities
$$\rho(\mathbf r)=\rho(r),$$
and you require only that $\rho(r)$ be an elementary function, then in general you can find the radial electric field as the integral
$$
E_r(r)=\frac{1}{\epsilon_0 r^2}\int_0^r\rho(r')r'^2\mathrm d r'
$$
but this will not in general be expressible as an elementary function. (For more details look up e.g. the Risch algorithm.) For the functions $\rho(r)$ that come up in practical, interesting problems, the integrals in question are generally doable but typically much harder than for an exponential. As you progress into electromagnetism, be prepared to work out much harder integrals.
I should note at this point that there is indeed an equivalent procedure, based not on integration but on differential equations, and it's correspondingly based on the differential form of Gauss's law,
$$
\frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2 E_r\right]=\nabla\cdot\mathbf E =\rho/\epsilon_0,
$$
or in terms of the potential
$$\nabla^2 V=-\rho/\epsilon_0.$$
This may or may not be more useful, depending on your tastes, but it should be immediately obvious that it's essentially the same as the integral version. There's no getting around the calculation.
