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 $x=\frac{3}{2}\sin\theta$.
 A: I tried to follow your steps. Here we are assuming that $0<x\leq 3/2$. Let $x=3/2\sin(t)$ then $$3\cos(t)=3\sqrt{1-(2x/3)^2}=\sqrt{9-4x^2}$$ and
\begin{align*}\int\frac{\sqrt{9-4x^2}}{x}dx&=\int\frac{3\cos(t)}{3/2\sin(t)}d(3/2\sin(t))\\&
=3\int\frac{\cos^2(t)}{\sin(t)}dt=3\int\frac{1-\sin^2(t)}{\sin(t)}dt\\&=3\int\frac{dt}{\sin(t)}-3\int \sin(t)dt
\\&=
3\ln\frac{1-\cos(t)}{\sin(t)}+3\cos(t)+C\\
\\&=
3\ln\frac{3-3\cos(t)}{3\sin(t)}+3\cos(t)+C\\
&=3\ln\frac{3-\sqrt{9-4x^2}}{2x}+\sqrt{9-4x^2}+C\\
&=3\ln\frac{3-\sqrt{9-4x^2}}{x}+\sqrt{9-4x^2}+C'.
\end{align*}
where $C'=C-3\ln(2)$ is an arbitrary constant.
P.S. We have that for $\sin(t)>0$,
\begin{align*}
\int\frac{dt}{\sin(t)}&=\int\frac{1+\cos(t)-\cos(t)}{\sin(t)}\,dt\\
&=\int\frac{(1+\cos(t))(1-\cos(t))}{\sin(t)(1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\
&=\int\frac{\sin(t)}{1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\
&=\ln(1-\cos(t))-\ln(\sin(t))+C=\ln\left(\frac{1-\cos(t)}{\sin(t)}\right)+C
\end{align*}
A: 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
$$\text{artanh}\sqrt{1-t^2}=\log\sqrt{\frac{1+\sqrt{1-t^2}}{1-\sqrt{1-t^2}}}=\log\left|\frac{1+\sqrt{1-t^2}}t\right|.$$
A: Here's what I did.
$$\int\frac{\sqrt{9-4x^2}}{x}dx \\
\begin{align}
& a=3\\
& u=2x\\
\text{since} \quad& u=a\sin\theta \\
& 2x=3\sin\theta \\
& 2dx=3\cos\theta; \quad x=\frac{3}{2}\sin\theta; \quad \sin\theta=\frac{2x}{3}; \quad \theta= \sin^1\frac{2x}{3}\\
& dx = \frac{3}{2}\cos\theta \\ \\
\sqrt{a^2-u^2}& =\sqrt{9-(2x)^2} \\
& = \sqrt{9-(3sin\theta)^2} \\
& = \sqrt{9-(9sin^2\theta)}\\
& = 3\cos\theta \\
\\
\text{Going back to the problem.}\\
\int\frac{\sqrt{9-4x^2}}{x}dx &=\int\frac{3\cos\theta}{\frac{3}{2}\sin\theta}(\frac{3}{2}\cos\theta d\theta) \\
&=3\int\frac{\cos^2\theta}{\sin\theta} \\
&=3\int\frac{1}{\sin\theta}-3\int\sin\theta \\
&=3\ln(\csc\theta-\cot\theta)+3\cos\theta \\
&=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+3\frac{\sqrt{9-4x^2}}{3} \\
&=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\
&=3\ln(\frac{3-\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\
&=3(\ln(3-\sqrt{9-4x^2}-\ln(2x))+\sqrt{9-4x^2} \\
&=3\ln(3-\sqrt{9-4x^2}-3\ln(2x)+\sqrt{9-4x^2} \\
\end{align}
$$
