Solve the integral $ \int{\frac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx $

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My attempt: \begin{align} \int{\frac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx &\\ &=\int{\frac{x+\sqrt{x^2+5x+6}}{x+\sqrt{x^2+5x+6}}}dx+\int{\frac{-\sqrt{x^2+5x+6}-\sqrt{x^2+5x+6}}{x+\sqrt{x^2-5x+6}}}dx \\ &=x-\int{\frac{\sqrt{x^2+5x+6}}{x+\sqrt{x^2+5x+6}}}dx-\int{\frac{\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}}dx \\ &=x-I_1-I_2 \end{align} I know how to solve the integral $I_1$ using Euler substitution, but I don't know how to solve the integral $I_2$.

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Hint:

$\int\dfrac{x-\sqrt{x^2-5x+6}}{x+\sqrt{x^2+5x+6}}~dx$

$=\int\dfrac{x-\sqrt{(x-3)(x-2)}}{\sqrt{(x+3)(x+2)}+x}~dx$

$=\int\dfrac{\left(x-\sqrt{(x-3)(x-2)}\right)\left(\sqrt{(x+3)(x+2)}-x\right)}{\left(\sqrt{(x+3)(x+2)}+x\right)\left(\sqrt{(x+3)(x+2)}-x\right)}~dx$

$=\int\dfrac{x\sqrt{(x+3)(x+2)}}{5x+6}~dx+\int\dfrac{x\sqrt{(x-3)(x-2)}}{5x+6}~dx-\int\dfrac{x^2}{5x+6}~dx-\int\dfrac{\sqrt{(x+3)(x+2)}\sqrt{(x-3)(x-2)}}{5x+6}~dx$