Be careful.

$\begin{aligned} A & = \int_1^4 x \ln \left(\frac {2x+1}{3x-2}\right)\ dx & \small \color{#3D99F6} \text{As }\ln \left(\frac ab\right) = \ln a - \ln b \\ & = {\color{#3D99F6}\int_1^4 x \ln \left(2x+1\right)\ dx} - \color{#D61F06} \int_1^4 x \ln \left(3x-2\right)\ dx \\ & = {\color{#3D99F6}I_1} - \color{#D61F06} I_2 \end{aligned}$

$\begin{aligned} I_1 & = \int_1^4 x\ln(2x+1)\ dx & \small \color{#3D99F6} \text{Let }u = 2x + 1 \implies x = \frac {u-1}2,\ dx = \frac {du}2 \\ & = \frac 14 \int_3^9 (u-1)\ln u \ du \\ & = {\color{#3D99F6}\frac 14 \int_3^9 u\ln u\ du} - \frac 14 \int_3^9 \ln u\ du & \small \color{#3D99F6} \text{Using integration by parts} \\ & = {\color{#3D99F6} \frac {u^2 \ln u}8 \bigg|_3^9 - \int_3^9 \frac u8\ du} - \frac {u\ln u - u}4 \bigg|_3^9 \\ & = \frac {81\ln 9-9\ln 3}8 - \frac {u^2}{16} \bigg|_3^9 - \frac {9\ln 9-9-3\ln 3+3}4 \\ & = \frac {123\ln 3}8 - 3 \end{aligned}$

$\begin{aligned} I_2 & = \int_1^4 x \ln \left(3x-2\right)\ dx & \small \color{#3D99F6} \text{Similar to finding }I_1 \\ & = \frac 19 \int_1^{10} (u+2)\ln u \ du \\ & = \frac 19 \left[\frac {u^2\ln u}2 - \frac {u^2}4 + 2u \ln u - 2u \right]_1^{10} \\ & = \frac {70\ln 10}9 - \frac {19}4 \end{aligned}$

Therefore, $A=I_1 - I_2 = \dfrac {123\ln 3}8 - 3 - \dfrac {70\ln 10}9 + \dfrac {19}4 = \dfrac {123\ln 9}{16} -\dfrac {70\ln 10}9 + \dfrac 74$ . Then $a+b+c = 9+7+4 = \boxed{20}$ .