Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$ 
I have almost the same solution where there's a $$3\ln|\csc\theta-\cot\theta|+3\cos\theta+C.$$ 
The problem is when the substitution comes in. I end up having 
$$3\ln\frac{|3-\sqrt{9-x^2}|}{2x}+\sqrt{9-x^2}$$ and when I tried to simplify it further, it resulted to $$3\ln\left|3-\sqrt{9-x^2}\right|-3\ln|2x|+\sqrt{9-x^2}.$$ I hope you could help me to tell where i did wrong. 
By the way, I set $a=3$ and $x=\frac{3}{2}\sin\theta$.

Here's what I did.
$$\int\frac{\sqrt{9-4x^2}}{x}dx \\
\begin{align}
& a=3\\
& u=2x\\
\text{since} \quad& u=a\sin\theta \\
& 2x=3\sin\theta \\
& 2dx=3\cos\theta; \quad x=\frac{3}{2}\sin\theta; \quad \sin\theta=\frac{2x}{3}; \quad \theta= \sin^1\frac{2x}{3}\\
& dx = \frac{3}{2}\cos\theta \\ \\
\sqrt{a^2-u^2}& =\sqrt{9-(2x)^2} \\
& = \sqrt{9-(3sin\theta)^2} \\
& = \sqrt{9-(9sin^2\theta)}\\
& = 3\cos\theta \\
\\
\text{Going back to the problem.}\\
\int\frac{\sqrt{9-4x^2}}{x}dx &=\int\frac{3\cos\theta}{\frac{3}{2}\sin\theta}(\frac{3}{2}\cos\theta d\theta) \\
&=3\int\frac{\cos^2\theta}{\sin\theta} \\
&=3\int\frac{1}{\sin\theta}-3\int\sin\theta \\
&=3\ln(\csc\theta-\cot\theta)+3\cos\theta \\
&=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+3\frac{\sqrt{9-4x^2}}{3} \\
&=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\
&=3\ln(\frac{3-\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\
&=3(\ln(3-\sqrt{9-4x^2}-\ln(2x))+\sqrt{9-4x^2} \\
&=3\ln(3-\sqrt{9-4x^2}-3\ln(2x)+\sqrt{9-4x^2} \\
\end{align}
$$
