<s>[INST] 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.
 [/INST] Check your book properly. For them to make the statement

...this is possible because $4c-b^2>0$

must mean that they have said somewhere above that they are considering this method subject to that restriction.
When such discussions are made, they're usually split into cases.
</s>