# Solve the Integral $\int \:\frac{3x+1}{\left(x^2-x-6\right)\sqrt{3x^2+4x+7}}$

$$\int \:\frac{3x+1}{\left(x^2-x-6\right)\sqrt{3x^2+4x+7}}$$

Can someone tell me what kind of substitution would work here and if this type of integral belongs to a certain group, that can be solved with a certain type of substitution, also a link to that type would be greatly appreciated.

P.S. There is no need to solve the problem for me, getting the right substitution and some explanation behind it is really plenty, thanks in advance.

Since $3x^2+4x+7=3\left[\left(x+\frac23\right)^2+\frac{17}9\right]$, a good substitution would be $$\frac{x+\frac23}{\sqrt{\frac{17}{9}}}=\frac{y-\frac1y}{2}.$$ It will reduce the problem to calculating antiderivative of a rational function of $y$ (think why).