Steady isothermal flow of an ideal gas

So I have a steady isothermal flow of an ideal gas through a smooth duct (no frictional losses) and need to compute the mass flow rate (per unit area) as a function of pressures at any two different arbitrary points, say 1 and 2. I have the following momentum equation in differential form:

$\rho vdv+dP=0$

where $v$ is the gas the flow velocity and $P$ is static pressure. The mass flow rate per unit cross section $G$, can be calculated by integrating this equation between points 1 and 2. This is where it gets confusing. I do the integration by two ways:

1) Use the ideal gas equation $P=\rho RT$ right away and restructure the momentum equation:

$vdv+RT\frac{dP}{P}=0$

, integrate it between points 1 and 2 and arrive at:

${v}_{1}^{2}-{v}_{2}^{2}+2RTln\frac{{P}_{1}}{{P}_{2}}=0$

Since the flow is steady, I can write $G={\rho}_{1}{v}_{1}={\rho}_{2}{v}_{2}$, again use the ideal gas law to write density in terms of gas pressure and finally arrive at the mass flow rate expression:

${G}^{2}=\frac{2ln\frac{{P}_{2}}{{P}_{1}}}{RT(\frac{1}{{P}_{1}^{2}}-\frac{1}{{P}_{2}^{2}})}$

2) In another way of integrating (which is mathematically correct), I start by multiplying the original momentum equation by $\rho $ to get

${\rho}^{2}vdv+\frac{1}{RT}PdP=0$

write ${\rho}^{2}v={G}^{2}/v$, integrate between points 1 and 2 to arrive at

${G}^{2}ln\frac{{v}_{2}}{{v}_{1}}=\frac{{P}_{1}^{2}-{P}_{2}^{2}}{2RT}$

Using the ideal gas law the velocity ratio can be written as the pressure ratio to finally arrive at the mass flow rate equation

${G}^{2}=\frac{{P}_{1}^{2}-{P}_{2}^{2}}{2RTln\frac{{P}_{1}}{{P}_{2}}}$

Both the expressions are dimensionaly sound and I know that the second expression is the correct one. My question is, whats wrong with first expression.