Integration.

$\begin{aligned} I & = \int \frac {dx}{(x-3)^2(x-4)^2} & \small \color{#3D99F6} \text{By partial fractions decomposition -- see below} \\ & = \int \left( \frac 2{x-3} + \frac 1{(x-3)^2} - \frac 2{x-4} + \frac 1{(x-4)^2} \right) dx \\ & = 2 \ln (x-3) - \frac 1{x-3} - 2 \ln (x-4) - \frac 1{x-4} + \color{#3D99F6}C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = 2 \ln \left(\frac {x-3}{x-4} \right) + 1 - \frac 1{x-3} - 1 - \frac 1{x-4} + C \\ & = \left(1 - \frac 1{x-3}\right) + \ln \left(\frac {x-4+1}{x-4} \right)^2 - \left(1 + \frac 1{x-4} \right) + C \\ & = \left(1 - \frac 1{x-3}\right) + \ln \left(1 + \frac 1{x-4} \right)^2 - \left(1 + \frac 1{x-4} \right) + C \end{aligned}$

$\implies p+q+r = 3+2+4 = \boxed{9}$

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Note: Assuming that (reference: Partial Fractions - Repeated Factors )

Let $\small \begin{aligned} \frac 1{(x-3)^2(x-4)^2} & = \frac A{x-3} + \frac B{(x-3)^2} +\frac C{x-4} + \frac D{(x-4)^2} \end{aligned}$ . Multiplying both sides by $\small (x-3)^2(x-4)^2$ and flipping sides.

$\small \begin{aligned} A(x-3)(x-4)^2 + B(x-4)^2 + C(x-3)^2(x-4) + D(x-3)^2 & = 1 & \small \color{#3D99F6} \text{Let }x = 3 \\ \implies B & = 1 & \small \color{#3D99F6} \text{Let }x = 4 \\ \implies D & = 1 & \small \color{#3D99F6} \text{Let }x = 2 \\ -4A+4B-2C + D & = 1 \\ \implies 2A + C & = 2 \quad ...(1) & \small \color{#3D99F6} \text{Let }x = 5 \\ 2A + B + 4C + 4D & = 1 \\ \implies 2A + 4C & = - 4 \quad ...(2) & \small \color{#3D99F6} (2)-(1) \\ \implies C & = - 2 & \small \color{#3D99F6} \text{Substitute in }(1) \\ \implies A & = 2 \end{aligned}$