<s>[INST] Integration trig substitution $\int \frac{dx}{x\sqrt{x^2 + 16}}$ $$\int \frac{dx}{x\sqrt{x^2 + 16}}$$
With some magic I get down to $$\frac{1}{4} \int\frac{1}{\sin\theta} d\theta$$
Now is where I am lost. How do I do this? I tried integration by parts but it doesn't work.
 [/INST] UPDATE 2 The first integral is not $\int \frac{dx}{x\sqrt{x^{2}-16}}$ but $\int 
\frac{dx}{x\sqrt{x^{2}+16}}$ (as noticed by  julien), because
\begin{eqnarray*}
I &=&\int \frac{dx}{x\sqrt{x^{2}+16}},\qquad x=\tan \theta ,dx=\sec
^{2}\theta d\theta  \\
&=&\int \frac{\sec ^{2}\theta }{\left( \tan \theta \right) 4\sec \theta }%
\,d\theta =\int \frac{\sec \theta }{4\tan \theta }\,d\theta  \\
&=&\int \frac{\sec \theta }{4\tan \theta }\,d\theta =\int \frac{1}{4\sin
\theta }\,d\theta \text{.}
\end{eqnarray*}

Use the Weierstrass substitution $$t=\tan\frac{\theta}{2}.$$
Then
$$\int \frac{1}{\sin \theta }\,d\theta =\int \frac{2}{\frac{2t}{1+t^{2}}
\left( 1+t^{2}\right) }\,dt=\int \frac{1}{t}\,dt=\ln \left\vert t\right\vert
+C=\ln \left\vert \tan \frac{\theta }{2}\right\vert +C.$$
Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form 
$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$
where $P,Q$  are polynomials in $\sin \theta,\cos \theta$ 
$$
\begin{equation*}
\tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt
\end{equation*},
$$
which converts the integrand into a rational function in $t$. We know from trigonometry that 
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{
\theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}
\frac{\theta }{2}}=\frac{2t}{1+t^2}.$$
Proof. A possible proof is the following one, which uses the double-angle formulas and the identity $\cos ^{2}\frac{\theta}{2}+\sin ^{2}\frac{\theta}{2}=1$:
$$
\begin{eqnarray*}
\cos \theta &=&\cos ^{2}\frac{\theta}{2}-\sin ^{2}\frac{\theta }{2}=\frac{\frac{\cos ^{2}
\frac{\theta}{2}-\sin ^{2}\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}{\frac{\cos ^{2}
\frac{\theta}{2}+\sin ^{2}\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}=\frac{1-\tan ^{2}
\frac{\theta}{2}}{1+\tan ^{2}\frac{\theta}{2}}, \\
&& \\
\sin \theta &=&2\sin \frac{\theta}{2}\cos \frac{\theta}{2}=\frac{\frac{2\sin \frac{\theta}{2}
\cos \frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}{\frac{\cos ^{2}\frac{\theta}{2}+\sin ^{2}
\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}=\frac{2\tan \frac{\theta}{2}}{1+\tan ^{2}
\frac{\theta}{2}}.
\end{eqnarray*}
$$

Another possible substitution is the Euler substitution
$$
\begin{equation*}
\sqrt{x^{2}+16}=t+x.
\end{equation*}
$$
Then
$$
\begin{eqnarray*}
I &=&\int \frac{dx}{x\sqrt{x^{2}+16}}=\int \frac{2}{t^{2}-16}\,dt \\
&=&\int \frac{1}{4\left( t-4\right) }-\frac{1}{4\left( t+4\right) }dt=\frac{1
}{4}\ln \left\vert \frac{t-4}{t+4}\right\vert +C \\
&=&\frac{1}{4}\ln \left\vert \frac{\sqrt{x^{2}+16}-x-4}{\sqrt{x^{2}+16}-x+4}
\right\vert +C.
\end{eqnarray*}
$$
</s>