Solving $\int{\frac{2x \ dx}{\sqrt{9x^2-6x-8}}}$ I solved the problem but my solution seems too simple to be the right one.
$$\int{\frac{2x \ dx}{\sqrt{9x^2-6x-8}}}$$
$$\frac{1}{3}\int{\frac{2x \ dx}{\sqrt{x^2-2x-\frac{8}{9}}}}$$
$u = x^2 - 2x - \frac{8}{9}$
$du = 2x -2 \ dx$
$2x \ dx = du + 2 = 3 du$
$$\frac{1}{3}\int{\frac{3\ du}{\sqrt{u}}} = \int{\frac{du}{\sqrt{u}}} = 2\sqrt{u} + K = 2\sqrt{x^2-2x-\frac{8}{9}} + K$$
Where did I go wrong?

Wait, how did you go from $$du = (2x-2) \, dx$$ to $$2x \, dx = 3 \, du?$$  That is not right.  You can't separate $(2x-2) \, dx$ into $2x \, dx - 2$ any more than you can write $(a+b)c = ac + b$.
