Evaluating $\displaystyle\int\frac{du}{\sqrt{-xu^{2}+yu+z}}$ This integral is just a step in a much longer problem for physics, but I am having some trouble with it.  

$$\int\frac{ \mathrm du}{\sqrt{-xu^{2}+yu+z}}$$

$x$, $y$ , and $z$ are constants
Also if the $-$ sign  in front of the first term is an issue (or any of the signs in front of any of the constants for that matter), I can easily fix that, as most of the function constants are just "dummy" placeholders for other terms.  
 A: 
Note the question says signs does not really matter here!



$$I=\int\frac{\mathrm du}{\sqrt{-xu^{2}+yu+z}}=\int\frac{\mathrm dx}{\sqrt{ax^{2}+bx+c}}$$

Using Euler Substitution
$$\sqrt{ax^{2}+bx+c}=x\sqrt a+t\iff x=\frac{c-t^2}{2t\sqrt a-b}=\iff \mathrm dx=-\frac{2(\sqrt a(c+t^2)-bt)}{(2t\sqrt a-b)^2}\mathrm dt$$

$$\begin{align}I&=\int\frac{\mathrm dx}{\sqrt{ax^{2}+bx+c}}\\
&=\int\frac{1}{\left({t+\sqrt a\dfrac{(c-t^2)}{2t\sqrt a-b}}\right)}\cdot\left(-\frac{2(\sqrt a(c+t^2)-bt)}{(2t\sqrt a-b)^2}\right)\mathrm dt\tag{1}\\
&=\int\frac{2}{b-2t\sqrt a}\mathrm dt\tag{2}\\
&=\frac{1}{\sqrt a}\ln\Big(b-2t\sqrt a\Big)+C\tag{3}\\
I&=\frac{1}{\sqrt a}\ln\left[b-2\sqrt a\Big(\sqrt{ax^2+bx+c}-x\sqrt{a}\Big)\right]+C\tag{4}\\
\end{align}$$

$$\int\frac{dx}{\sqrt{ax^{2}+bx+c}}=\frac{1}{\sqrt a}\ln\left[b-2\sqrt a\Big(\sqrt{ax^2+bx+c}-x\sqrt{a}\Big)\right]+C$$


$\text{Explanations}\\ 1.\text{ Euler Substitution}\\
2. \text{ Magic }\\
3. \text{ Integrating (2) }\\
4. \text{ Putting Everything back }$
A: Hint:
Complete the square under the radical so that its argument looks something like $(x+a)^2 + c$ for some $a, c \in \mathbb{R}$.
At that point, you can finish up with a trig substitution.
