<s>[INST] Help with a trig-substitution integral I'm in the chapter of trigonometric substitution for integrating different functions. I'm having a bit of trouble even starting this homework question:
$$\int \frac{(x^2+3x+4)\,dx} {\sqrt{x^2-4x}}$$
 [/INST] 
Make the substitution suggested by completing the square, $\sqrt{x^2-4x}=\sqrt{(x-2)^2-4}$, that is, $y=x-2$. Then, by the easily constructed triangle, we have
\begin{align}
\frac{y}{2}
 & = \sec\theta\\
\frac{\sqrt{y^2-4}}{2}
 & = \tan\theta\\
x^2+3x+4
 & =(y+2)^2+3(y+2)+4\\
 & = y^2+7y+14\\
\frac{dy}{2}
 & = \sec\theta\tan\theta\ d\theta. 
\end{align}
The integral becomes 
\begin{align}
\int \frac{(x^2+3x+4)} {\sqrt{x^2-4x}}dx
 & = \int\frac{y^2+7y+14}{\sqrt{y^2-4}}dy\\
 & = \int\frac{4\sec^2\theta+14\sec\theta+14}{2\tan\theta}2\sec\theta\tan\theta\ d\theta\\
 & = \int4\sec^3\theta+14\sec^2\theta+14\sec\theta\ d\theta.
\end{align}
Andre Nicolas suggests in another response that the powers of secant are unsavory, and that a hyperbolic trig sub would be more apt. (Claude Leibovici explores this.) 
(I vaguely recall methods for the $\sec^3\theta$ integrand in Stewart's Calculus. The $\sec\theta$ is in most trig integral lists, and perhaps the $\sec^2\theta$ is amenable to some fortuitous trig identity.) 
Here are some resources on those: 
http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_secant
http://en.wikipedia.org/wiki/Integral_of_the_secant_function
http://en.wikipedia.org/wiki/Integral_of_secant_cubed
</s>