How to compute the integral $ \int\frac{1}{x\sqrt{x^2 +3x}}dx$ Given a problem :
$$ \int\frac{1}{x\sqrt{x^2 +3x}}dx, $$
what is the best solution for this?
I am thinking about solving this problem by using :
$$ u = x+3;\qquad x = u-3; $$
So that we get :
$ \int\frac{1}{x\sqrt{x}\sqrt{x+3}} dx$, then $ \int\frac{1}{(u-3)^{3/2}(u)^{1/2}} du$,
then 
$ \int (u)^{-1/2} (u-3)^{-3/2}  du$.
Am I right so far? or is there a better method? Thanks.
 A: I have not done an integral this cumbersome in quite a while. The technique that immediately stands out to me is trigonometric substitution. While I have performed the integration correctly on paper, I would appreciate someone being on the hunt for typesetting errors. Buckle seatbelt...
We have 
$$\int \frac{1}{x\sqrt{x^2+3x}} dx.$$
Now we complete the square on the radicand to get
$$\int \frac{1}{x\sqrt{x^2+3x}} dx=\int \frac{1}{x\sqrt{\left( x+\frac{3}{2} \right)^2-\frac{9}{4}}} dx.$$ 
For our substitution, let
\begin{align*}
&x+\frac{3}{2}=\frac{3}{2}\sec(\theta) \\
\Rightarrow & x=\frac{3}{2}(\sec(\theta)-1) \\
\Rightarrow & dx=\frac{3}{2}\sec(\theta)\tan(\theta)d \theta. \\
&\left( x+\frac{3}{2} \right)^2=\frac{9}{4}\sec^2(\theta) \\
\Rightarrow &\left( x+\frac{3}{2} \right)^2-\frac{9}{4}=\frac{9}{4}\sec^2(\theta)-\frac{9}{4} \\
& \qquad \qquad \qquad \quad= \frac{9}{4}\tan^2(\theta).
\end{align*}
We now make our substitutions and then integrate with respect to $\theta$.
\begin{align*}
\int \frac{1}{x\sqrt{\left( x+\frac{3}{2} \right)^2-\frac{9}{4}}} dx &=\int\frac{\frac{3}{2}\sec(\theta)\tan(\theta) d \theta}{\frac{3}{2}(\sec(\theta)-1)\sqrt{\frac{9}{4}\tan(\theta)}} \\
&=\frac{2}{3}\int \frac{\sec(\theta)\tan(\theta) d \theta}{\tan(\theta)(\sec(\theta)-1)} \\
&=\frac{2}{3}\int\frac{\sec(\theta)d \theta}{\sec(\theta)-1} \\
&=... \\
&=\frac{2}{3}\int \left( \csc^2(\theta)+\csc(\theta)\cot(\theta) \right) d\theta \\
&=\frac{2}{3}\left( -\cot(\theta)-\csc(\theta) \right) +c.
\end{align*}
Finally we back substitute (a right triangle helps). We started with the substitution
$$x+\frac{3}{2}=\frac{3}{2}\sec(\theta) \Rightarrow \sec(\theta)=\frac{2x+3}{3}.$$
If we form a right triangle with $3$ adjacent to $\theta$, and $2x+3$ on the hypoteneuse (this follows as $\sec(\theta)=\frac{2x+3}{3}$), then we find that the opposite side is $2\sqrt{x^2+3x}.$
Thus the back substitution becomes (reading straight from the right triangle),
\begin{align*}
&\frac{2}{3}\left( -\cot(\theta)-\csc(\theta) \right) +c \\
=&\frac{2}{3}\left( -\frac{3}{2\sqrt{x^2+3x}} -\frac{2x+3}{\sqrt{x^2+3x}}\right) +c \\
=& ... \\
=& -\frac{2\sqrt{x^2+3x}}{3x}+c.
\end{align*}
I will be happy to include the derivations of the trigonometric moves, and this final simplification upon request where I inserted "..." into the process. Transforming $$\frac{\sec(\theta)}{\sec(\theta)-1}=\csc^2(\theta)+\csc(\theta)\cot(\theta)$$ was not what I would call trivial (took me 6 steps or so). A picture of the right triangle for the back substitution would also enhance this, and I could also add that later unless someone else feels more motivated right now.
A: For $x>0$, $$\int\frac{1}{x\sqrt{x^2 +3x}}dx=\int\frac{1}{x^2\sqrt{1 +\frac{3}{x}}}dx = -\frac{1}{3}\int\frac{du}{\sqrt{u}}=-\frac{2}{3}\sqrt{u}=-\frac{2}{3}\sqrt{1+\frac{3}{x}} +C $$ where $$u=1+\frac{3}{x}$$
A: Hint: Let $y^2= 1+\dfrac{3}{x}$, i.e. $x=\dfrac{3}{y^2-1}$.

 Then $dx=-\dfrac{6ydy}{(y^2-1)^2}$.

A: The Maple code with(Student[Calculus1]):IntTutor(1/(x*sqrt(x^2+3*x)), x); 
produces that step by step with explanations.
