Let $\dfrac yx = v$. Then, $\dfrac{\mathrm dy}{\mathrm dx} = v+x\dfrac{\mathrm dv}{\mathrm dx}$
$$\begin{array}{rcl} (4y+3x) \ \mathrm dy + (y-2x) \ \mathrm dx &=& 0 \\ \dfrac{\mathrm dy}{\mathrm dx} &=& \dfrac{2-\frac yx}{3+4\frac yx} \\ v+x\dfrac{\mathrm dv}{\mathrm dx} &=& \dfrac{2-v}{3+4v} \\ x\dfrac{\mathrm dv}{\mathrm dx} &=& \dfrac{2-4v-4v^2}{3+4v} \\ \displaystyle \int \dfrac{(4v+3)\ \mathrm dv}{-4v^2-4v+2} &=& \displaystyle \int \dfrac{\mathrm dx}{x} \\ \end{array}$$
I am not able to solve this integral. Please help.
To find $\displaystyle \int \dfrac{(4v+3)\ \mathrm dv}{-4v^2-4v+2}$:
$$\begin{array}{rcl} 4v+3 &=& A(-8v-4) + B \\ A &=& -\dfrac{1}{2} \\ B &=& 1 \end{array}$$
$$\begin{array}{cl} &\displaystyle \int \dfrac{(4v+3)\ \mathrm dv}{-4v^2-4v+2} \\ =& \displaystyle -\dfrac12 \int \dfrac{(-8v-4)\ \mathrm dv}{-4v^2-4v+2} + \int \dfrac{\mathrm dv}{-4v^2-4v+2} \\ =& \displaystyle -\dfrac12 \int \dfrac{\mathrm d(-4v^2-4v+2)}{-4v^2-4v+2} + \int \dfrac{\mathrm dv}{-4(v+0.5)^2+3} \\ =& \displaystyle -\dfrac12 \ln\left|(-4v^2-4v+2)\right| - \dfrac14 \int \dfrac{\mathrm dv}{(v+0.5)^2-0.75} \\ =& \displaystyle -\dfrac12 \ln\left|(-4v^2-4v+2)\right| - \dfrac1{4\sqrt3} \int \left(\dfrac{1}{v+0.5-\sqrt{0.75}} - \dfrac{1}{v+0.5+\sqrt{0.75}}\right) \ \mathrm dv \\ =& \displaystyle -\dfrac12 \ln\left|(-4v^2-4v+2)\right| - \dfrac1{4\sqrt3} \left(\ln \left|(v+0.5-\sqrt{0.75})\right| - \ln \left|(v+0.5+\sqrt{0.75})\right|\right) + C\\ \end{array}$$