Directly calculate electric potential due to a spherical volume charge Q:
Say we have a uniformly charged sphere with radius $R$, total charge $Q$. Now we need to find out the electric potential inside the sphere, at a distance r from the centre. I understand that we can do this by using 
$$\mathbf{E}=-\mathbf{\nabla}V. \tag{1}$$ 
But is there a way to do this by using: 
$$V(r)=\int {\frac{\rho(r)}{r}d{\tau}} \tag{2}$$ 
to directly calculate the potential and then get the electric field from integration?
I have no idea how to do this, could someone please give me a hint on how to get started?
 A: Generically the potential is given by the integral,
$$ \Phi(x) = \int \rho(x') \frac{1}{\vert x - x' \vert} d\tau' ,$$
which is based on Coulomb's law of Electrostatics. 
It is possible to use the spherical symmetry of $\rho$ to show that $\Phi(\Omega x ) = \Phi(x) $ for any proper rotation $\Omega$. This means we can set $x$ to be on the $z$-axis without loss of generality. 
Now we write the integral in more detail ultimately using spherical coordinates for the integration.
$$ \Phi(x) =  \int \rho(x') \frac{1}{\sqrt{ x^2 + x'^2 - 2 x \cdot x'}} d\tau',$$
$$ = \int_0^{2\pi} \int_0^\pi \int_0^R \rho(r) \frac{r^2 \sin(\theta) dr d\theta d\phi }{\sqrt{ z^2 + r^2 - 2 zr\cos(\theta)}}  .$$
The integrand is independent of $\phi$ so we can perform that integration automatically getting a factor of $2\pi$ out front. We can then handle the term in the square root by performing a change of integration variable for $\theta$. Let $y=z^2+r^2-2zr \cos(\theta)$ then $dy =2zr \sin(\theta) d\theta$.
$$ \Phi(x) = 2\pi \int_0^\pi \int_0^R \rho(r) \frac{r^2 \sin(\theta) dr d\theta  }{\sqrt{ z^2 + r^2 - 2 zr\cos(\theta)}} $$
$$ = 2\pi  \int_0^R \left[ \int_0^\pi \rho(r) \frac{r^2 \sin(\theta)  d\theta  }{\sqrt{ z^2 + r^2 - 2 zr\cos(\theta)}} \right]dr $$
$$ = 2\pi  \int_0^R \left[ \int_{(z-r)^2}^{(z+r)^2} \rho(r) \frac{r^2 \frac{1}{2zr} dy }{\sqrt{y}} \right]dr $$
$$ = \frac{2\pi}{z}  \int_0^R \rho(r) r \left[   \sqrt{y} \right]_{(z-r)^2}^{(z+r)^2} dr $$
$$ = \frac{2\pi}{z}  \int_0^R \rho(r) r \left[   (z+r) - \left| z-r \right| \right]dr $$
If we are outside the charge distribution then we can write,
$$ \Phi = \frac{4\pi}{z}  \int_0^R \rho(r) r^2  dr $$
In the case of a uniform charge distribution we would get,
$$ \Phi = \frac{4\pi R^3 \rho }{3z} = \frac{q}{z} \Rightarrow \boxed{ \Phi(x) = \frac{q}{\vert x \vert} }$$

There are some concerns in the comments that I didn't calculate the potential inside the sphere. I wanted to leave this step up to the OP, but since he has other ways of computing the correct answer I do not see the harm in reproducing it here.
We proceed the same as before except now $0 < z < R$ whereas we had $z>R$. We will eventually arrive at,
which will be broken into to integrations : one for $z<r$ and the other for $r<z<R$. We will then use our knowledge of whether $z$ is larger or smaller than $r$ to evaluate the abolute value $\vert z-r \vert$. 
$$\Phi(z) = \frac{2\pi}{z}  \left( \int_0^z \rho(r) r \left[   (z+r) - \left| z-r \right| \right]dr + \int_z^R \rho(r) r \left[   (z+r) - \left| z-r \right| \right]dr \right), $$
$$\Phi(z) = \frac{2\pi}{z}  \left( \int_0^z \rho(r) r \left[   (z+r) - ( z-r) \right]dr + \int_z^R \rho(r) r \left[   (z+r) - ( r-z) \right]dr \right), $$
$$ \Phi(z) = \frac{2\pi}{z}  \left( \int_0^z \rho(r) 2r^2dr + \int_z^R \rho(r) 2rz dr \right), $$
$$ \Phi(z) = \frac{2\pi}{z}  \left( \frac{1}{2 \pi} \int_0^z \rho(r) 4\pi r^2dr + \int_z^R \rho(r) 2rz dr \right), $$
The left hand integral is just the enclosed charge less than $z$. So we can rewrite this as, 
$$ \Phi(z) = \frac{Q_{enc}(z)}{z}   + 4\pi \int_z^R \rho(r) r dr . $$
Notice that the upper bound is irrelevant in the right hand integral because all it does is contribute and additive constant to the potential. It is only the charges located at $r \leq z$ which contribute to the actual physics. We could obtain the electric field from this expression by using the fundamental theorem of calculus to differentiate the integrals with respect to $z$. 
An interesting special case of this expression is when $\rho (r) $ is constant. In that case we get,
$$ \Phi(z) = \frac{\rho 4\pi z^3/3}{z}   - 2\pi z^2 \rho + (constants) , $$
$$ \Phi(z) = - \frac{2\pi}{3} \rho z^2  + (constants) . $$
This predicts that the Electric Field within the sphere grows linearly with $z$ points away from the center if the charge distribution is positive. 
A: It is not clear whether the formula (2) in the question, is correct.
Let's calculate the field $E$ by Gauss' integral law, taking in consideration the spherical symmetry and the fact that the field is parallel to the normal of any sphericl surface concentric with the sphere of radius $R$
$$\int_0^r \frac {\rho}{\epsilon_0} (4\pi r'^2 dr') = \frac {4\pi r^2}{\epsilon_0} E(r). \tag{i}$$
Thus, one gets
$$E(r) = \frac {\rho}{\epsilon_0} \frac {r}{3}, \ \ \ \text {or,} \ \ \ E(r)= \frac {Q(r)}{4\pi \epsilon_0 r^2}, \tag{ii}$$
where $Q(r)$ means the charge encapsulated inside the spherical volume of radius $r$. Now we can calculate $V(r)$ and see if the formula (1) in the question is correct.
$$V(r) = \int_0^r E \ dr' = \frac {\rho \ r^2}{6\epsilon_0}, \ \ \ \text {or,} \ \ \ V(r) = \frac {Q(r)}{8\pi \epsilon_0 r}. \tag{iii}$$
Now, let's calculate $V(r)$ with the formula (2) in the question, in which let's introduce the multiplicative constant $1/(12\pi\epsilon_0)$.
$$V(r) = \frac {1}{12\pi\epsilon_0}\int_0^r \frac {\rho}{r'} \ d\tau = \int_0^r \frac {\rho}{3\epsilon_0 r'} \ (r'^2 dr') = \frac {\rho r^2}{6\epsilon_0}, \ \ \ \text {or,} \ \ \ V(r) = \frac {Q(r)}{8\pi \epsilon_0 r}. \tag{iv}$$
Now by comparing with $\text {(iii)}$, the formula (2) in the question is confirmed.
About your last question, how to calculate the electrical field, you can either use formula $\text {(ii)}$, apply your formula (1) to the before-last expression in $\text {(iii)}$ or in $\text {(iv)}$. Again, due to the spherical symmetry of the problem the gradient of the potential is along the direction of the field, simply $E(r) = \frac {dV(r)}{dr}$. You can check that you get again the formulas $\text {(ii)}$.
A: For a different way using Laplace's Equation look at
http://hyperphysics.phyastr.gsu.edu/hbase/electric/laplace.html
Hmmmm: Not sure, but the link doesn't seem to work: alternatively search for
LaPlace's and Poisson's Equations hyperphysics.phy-astr.gsu.edu
