How should the integral be used when trying to find the heat rate in a cylinder which has two temperatures? The problem is as follows:

A coaxial cylinder has an interior radius $r_{1}$ and a temperature
  $T_{1}$ and a external radius $r_{2}$ and a temperature $T_{2}$ and
  has a height $h$. Assuming the thermal conductivity is $k$. Find the
  heat rate which is flowing radially.

The alternatives given are as follows:
$\begin{array}{ll}
1.&2\pi kh\frac{T_2-T_1}{\ln\frac{r_1}{r_2}}\\
2.&2\pi kh^2\frac{T_1-T_2}{r_2-r_1}\\
3.&2\pi kh\frac{T_1-T_2}{\ln\frac{r_2}{r_1}}\\
4.&2\pi kh^2\frac{T_2-T_1}{r_2-r_1}\\
5.&2\pi kh\frac{T_2-T_1}{\ln{r_2}{r_1}}\\
\end{array}$
I'm not exactly how to handle the integration for this problem. I'm assuming that the intended principle for this problem will be given by:
The Fourier's law is:
$q=-k A \frac{dT}{dx}$
Hence the area in the coaxial cylinder will be:
$A=(\pi r_2^2-\pi r_1^2)=\pi(r_2^2-r_1^2)$
$q=-k \pi(r_2^2-r_1^2) \frac{dT}{dx}$
I'm assuming that the integration is between $T_1$ and $T_2$ but I don't know how to assemble the Fourier Biot equation to adequately integrate it. Can someone help me here?. Since I dont know how this process is happening. Can someone include some sort of sketch or diagram to see how is the direction of the heat flowing?. As I don't understand how is the heat being integrated here. 
 A: In Conduction in the Cylindrical Geometry, your question (apart from them using $L$ for the cylinder (which they refer to specifically as a pipe) length instead of $h$) is explained and answered in quite of bit of detail (including several diagrams), with their result near the bottom of page $4$ being

$Q = \ldots = 2\pi Lk\frac{T_1 - T_2}{\ln(r_2/r_1)}$

This corresponds to your alternative #$3$. I'll outline what's written there.
As indicated, they use a balance of heat in & heat out approach using cylindrical shells, going the entire length $h$, of inner radius $r$ and outer radius $r + \Delta r$.
Have $Q(r)$ be the radial heat flow within the cylinder wall at a radial distance of $r$. Thus, the heat flow into the cylindrical shell is $Q(r)$ and the outward heat flow on the other side is $Q(r + \Delta r)$. At steady state, you have
$$\begin{equation}\begin{aligned}
Q(r + \Delta r) & = Q(r) \\
Q(r + \Delta r) - Q(r) & = 0 \\
\frac{Q(r + \Delta r) - Q(r)}{\Delta r} & = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Since $Q(r)$ is differentiable, taking the limit as $\Delta r \to 0$ gives the very simple differential equation of
$$\frac{dQ(r)}{dr} = 0 \tag{2}\label{eq2A}$$
Integrating this leads to
$$Q(r) = C \tag{3}\label{eq3A}$$
for some constant $C$, i.e., it's independent of the radial location, i.e, $r$.
Next, note
$$Q(r) = q_rA \tag{4}\label{eq4A}$$
where $q_r$ is the heat flux in the radial direction and
$$A = 2\pi rh \tag{5}\label{eq5A}$$
is the area of the cylindrical surface normal to the $r$-direction along the cylinder of length $h$. Next, you have Fourier's law which states
$$q_r = -k\frac{dT}{dr} \tag{6}\label{eq6A}$$
Substituting \eqref{eq6A} into \eqref{eq4A}, plus using \eqref{eq3A} and \eqref{eq5A}, gives
$$\begin{equation}\begin{aligned}
\left(-k\frac{dT}{dr}\right)\left(2\pi rh\right) & = C \\
(-2k\pi h)r\frac{dT}{dr} & = C \\
r\frac{dT}{dr} & = \frac{C}{-2k\pi h} \\
r\left(\frac{dT}{dr}\right) & = C_1
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
where $C_1$ is a constant to be determined later using the boundary conditions. Note \eqref{eq7A} is a separable equation, so using separation of variables to solve it gives
$$T(r) = C_1\ln r + C_2 \tag{8}\label{eq8A}$$
where $C_2$ is another constant that can be solved for later, and is determined in the linked paper, but which I don't show here as it's not used in, and doesn't affect, the result.
Using the boundary conditions of $T(r_1) = T_1$ and $T(r_2) = T_2$ results in
$$C_1 = \frac{T_1 - T_2}{\ln(r_1/r_2)} \tag{9}\label{eq9A}$$
Using the above equations of \eqref{eq4A}, \eqref{eq6A}, \eqref{eq7A} and \eqref{eq8A},you now get
$$\begin{equation}\begin{aligned}
Q & = q_rA \\
& = \left(-k\frac{dT}{dr}\right)2\pi rh \\
& = \left(-kC_1\right)2\pi h \\
& = -2\pi hk\frac{T_1 - T_2}{\ln(r_1/r_2)} \\
& = 2\pi kh\left(\frac{T_1 - T_2}{\ln(r_2/r_1)}\right)
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$
This is basically the equation I quoted initially, except with their $L$ replaced with $h$.
