<s>[INST] Quadratic Expressions: Advanced techniques of Integration $$\int \frac{x}{\sqrt{5+12x-9x^2}}\,dx$$
After two steps I arrive at $\displaystyle{ \int \frac{x}{\sqrt{9-(3x-2)^2}}}\,dx$
Using trigonometric substitution, we have a triangle with a cosine of $\theta$ of $\displaystyle{\sqrt{9-(3x-2)^2}\over 3}$ and a $\sin(\theta$) of $(3x-2)\over (3)$. 
The final two steps of solving give
$\displaystyle {\frac{-1}{3}\cos(\theta) + \frac{2}{9}\theta +C}$
Therefore
-$\displaystyle{\sqrt{9-(3x-2)^2}\over 9}$ 
Wolfram alpha says that there should be a positive sign, i.e. -$\displaystyle{\sqrt{9+(3x-2)^2}\over 9}$
How can that be possible given the value for cosine of theta? 
 [/INST] setting $$\sqrt{5+12x-9x^2}=xt+\sqrt{5}$$ we get $$x=\frac{12-2t\sqrt{5}}{t^2+9}$$ and we get $$dx=\frac{2(-9\sqrt{5}-12t+\sqrt{5}t^2)}{(9+t^2)^2}dt$$ and our integral is rational.
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