Integrating pressure with respect to time I am trying to work through the math derivation presented in a paper about gas flowing through rock due to a pressure differential across the length of rock.  This is my first post so forgive me if I am going about this the wrong way.  My question is about how to perform the math between two steps in the derivation in the paper, but I feel that the steps leading up to the steps in question are necessary since and idea of what the pressures are in relation to the system being evaluated is useful in the math derivation.
To begin, the volumetric isothermal flow rate of nitrogen, which behaves as an ideal gas, from a storage tank at pressure $P_0$ is: 
\begin{equation}
q_o(t)=\frac{M}{\rho_o(t)}\frac{-dn}{dt}=\frac{-MV_t}{\rho_o(t)RT}\frac{dP_o}{dt}  
\end{equation}
But density is given by: $$\rho_o(t)=\frac{MP_o(t)}{RT}$$ 
Therefore, $$q_o(t)=\frac{-V_t}{P_o(t)}\frac{dP_o}{dt}$$
Assume for the moment that at any instant in time the mass velocity throughout the length of the core is constant (this is not rigorously true).  As nitrogen flows through a rock core, it expands, such that
$$q(x,t)=\frac{q_o(t)P_o(t)}{p(x,t)}=\frac{-V_t}{p(x,t)}\frac{dP_o}{dt}$$
Klinkenberg's relationship for permeability to gas, expressed as a point function of both time and position, is:
$$k(x,t)=k_l\left(1+\frac{b}{p(x,t)+Pa}\right)$$
Darcy's relation for fluid flow in a porous medium in 1D is:
$$q = \frac{-kA}{\mu}\frac{dP}{dx}$$
Substituting our volumetric relation and Klinkenberg's relationship into Darcy's equation for one-dimensional flow yields:
$$\frac{-V_tP'_o(t)}{p(x,t)}=\frac{-k_lA(1+b/p(x,t))}{\mu}\frac{\partial{p(x,t)}}{\partial{x}}$$
If this equation is integrated with respect to length and divided by $1/2(P_l-P_o)$, it becomes:
$$\frac{-2V_t\mu P'_o(t)L}{k_lA(P_o-P_l)}=P_l+P_o+2b$$
Pressures in all equations above have been absolute pressures, expressed in atmospheres and permeability in darcies.  If we now switch to gauge pressure (psig), and express permeability in millidarcies, our equation becomes (since $P_l = 0$ psig):
$$\frac{-V_tP'_o(t)}{P_o(t)}=\frac{k_lA}{2000*14.696 \mu L}(P_o(t)+2P_a+2b)$$
set
$$m=\frac{k_lA}{29,390 \mu L}$$  and $$i=2(P_a+b)m$$
we have an equation of a straight line with intercept $i$ and slope $m$:
$$\frac{-V_tP'_o(t)}{P_o(t)}=i+mP_o(t)$$
Now here is the part that I would like a math explanation.  The paper states that our equation can be rearranged and integrated with respect to time to give:
$$\frac{V_t}{i/m} \ln \frac{P_1(P_2+i/m)}{P_2(P_1+i/m)}=m(t_2-t_1)$$
How exactly was this step done?  Thanks in advance for any help.
 A: Rearrange the equation to read
$$-V_t\frac{P_0'}{P_0^2} - \frac{i}{P_0} = m$$
Then rewrite as
$$ \frac{d}{dt} \frac{1}{P_0} - \frac{i}{V_t} \frac{1}{P_0} = \frac{m}{V_t}$$
This may be re-expressed using an integrating factor:
$$\frac{d}{dt} \left [e^{-i t/V_t} \frac{1}{P_0} \right ] = \frac{m}{V_t} \, e^{-i t/V_t}$$
Integrating, we get
$$\frac{1}{P_0(t)} = C \, e^{i t/V_t} - \frac{m}{i}$$
where $C$ is a constant of integration.  Rearranging and taking logs, we get
$$\log{\left (\frac{1}{P_0(t)}+\frac{m}{i} \right )} - \frac{i}{V_t} t = \log{C}$$
This means, for times $t=t_1$ and $t=t_2$ we have
$$\log{\left (\frac{1}{P_0(t_1)}+\frac{m}{i} \right )} - \frac{i}{V_t} t_1 = \log{\left (\frac{1}{P_0(t_2)}+\frac{m}{i} \right )} - \frac{i}{V_t} t_2$$
Set $P_1 = P_0(t_1)$ etc., and the rest is algebra.
A: @Ron -- I will have to study your answer a bit more. I really appreciate your help. 
After rearranging the equation to $$\frac{-V_t}{P_o(t)(P_o(t)+i/m)}\frac{dP_o}{dt}=m$$
I ended up using the idea of partial fractions; setting $V_t = C$, $(i/m) =D$ and $P_o(t)=x$ So that I have $$\frac{-C}{x(x+D)}dx=mdt$$ Rearraging the LHS $$\frac{-C}{x(x+D)}=\frac{A}{x}+\frac{B}{x+D}$$  Making $$-C=A(x+D)+B(x)$$  If $x=0$, then $A=-C/D$
If $x=-D$, then $B=C/D$
Therefore we have $$\frac{-C}{x(x+D)}=\frac{-C/D}{x}+\frac{C/D}{x+D}$$ 
$$=\frac{C}{D} \left( \frac{-1}{x}+\frac{1}{x+D} \right)$$
After integrating we have $$\frac{C}{D} \left [\ln(x_1)-\ln(x_2)+\ln(x_2+D)-\ln(x_1+D) \right]$$  From rules of natural logs  we can say $$\frac{C}{D} \ln \frac{x_1(x_2+D)}{x_2(x_1+D)}$$
And then substitue back our original variables...
