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 $x=\frac{3}{2}\sin\theta$.

It is easy to mess up with the constants in such integrals. But we can temporarily ignore them because a simple scaling of the variable and of the integrand can bring us to
$$I:=\int\frac{\sqrt{1-t^2}}tdt.$$
This calls for the substitution $1-t^2=u^2$, giving $dt/t=-u\,du/2t^2$ and
$$I\propto\int\frac{u^2}{u^2-1}du=\int\frac{u^2-1+1}{u^2-1}du=u-\text{artanh u}=\sqrt{1-t^2}-\text{artanh}\sqrt{1-t^2}.$$

Now back to the original problem, we now know that the solution will be of the form
$$\int\frac{\sqrt{9-4x^2}}xdx=\lambda\left(\sqrt{1-\frac49x^2}-\text{artanh}\sqrt{1-\frac49x^2}\right).$$
Differentiating, we get
$$\frac{\sqrt{9-4x^2}}x=-\lambda\frac{\dfrac49x}{\sqrt{1-\dfrac49x^2}}\left(1-\frac1{1-\left(1-\dfrac49x^2\right)}\right)=\lambda\frac{\sqrt{1-\dfrac49x^2}}x$$ and $\lambda=3$.
The final solution is
$$\sqrt{9-4x^2}-3\text{ artanh}\frac{\sqrt{9-4x^2}}3.$$

Note:
We can use the following transformation
$$\text{artanh}\sqrt{1-t^2}=\log\sqrt{\frac{1+\sqrt{1-t^2}}{1-\sqrt{1-t^2}}}=\log\left|\frac{1+\sqrt{1-t^2}}t\right|.$$
