Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-substitution:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2+\alpha^2
$$
where $u=x+\frac{b}{2}$ and $\alpha=\frac{1}{2}\sqrt{4c-b^2}$. The book says: "... this  is possible because $4c-b^2>0$."
Eagerly I tried an example, using the quadratic $x^2-8x+1$.
Then let $a=1, b=-8, c=1$ and:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2 + \alpha^2
$$
where $u = x+b/2 = x-4$ but we run into a problem: 
$$\alpha=\frac{1}{2}\sqrt{4c-b^2} =  \frac{1}{2}\sqrt{(4)(1) - (-8)^2} = \frac{1}{2}\sqrt{4-64}$$
So $\alpha$ doesn't satisfy $4c-b^2>0$. Maybe I'm missing something obvious? Or is the book missing a caveat that this method doesn't always work. Because in the book they make it sound like "... this  is possible because $4c-b^2>0$." is always true.

The answer is that this method doesn't always work. In this particular case, we can use the following: Since
$$x^2-8x+1=0 \quad \Leftrightarrow \quad x=\frac{8\pm \sqrt{60}}{2}=4\pm \sqrt{15}$$
this polynomial can be factored as $$\big(x-(4+\sqrt{15})\big)\big(x-(4-\sqrt{15})\big)$$
use partial fraction like this
$$\frac{1}{x^2-8x+1}=\frac{1}{(x-4-\sqrt{15})(x-4+\sqrt{15})}=\frac{A}{x-4-\sqrt{15}}+\frac{B}{x-4+\sqrt{15}}$$
and proceed. Or maybe, since $x^2-8x+1=(x-4)^2-15$ just set $x-4=\sqrt{15}\sec \theta$ (trigonometric substitution).
