Integrating quadratics in denominator

Question

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.

Answer 1

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).

Answer 2

For integrating the Quadratics in the denominator, there are two cases,

$b^2-4c>0:$

$x^2+bx+c= (x+\frac{b}{2})^2-\frac{b^2-4c}{4} = u^2 - \beta^2 \Rightarrow $ This will give you: $\frac{1}{2 \beta} \ln \frac{u- \beta} {u+ \beta}; (\beta> 0)$

$4c-b^2>0:$

$x^2+bx+c= (x+\frac{b}{2})^2+\frac{4c-b^2}{4} = u^2 + \alpha^2 \Rightarrow $ This will give you: $\frac{1}{\alpha} \arctan \frac{u} {\alpha}; (\alpha >0)$

Your denominator x^2-8x+1 belongs to FIRST case.

Answer 3

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.