How to integrate $\frac{10}{4x^2-24x+61}$? 
Show that $\displaystyle\int_3^{5.5}\dfrac{10}{4x^2-24x+61}\mathrm dx=\dfrac\pi4.$

I've completed the square, and now have:
$10 \int \dfrac{1}{4(x-3)^2+25}dx$
Using common results, I know it should be:
$10 \cdot z \cdot  \arctan\dfrac{2(x-3)}{5}$
I know $z=1/10$, but how do I get $1/10$ from those numbers?

Since we generally compute these integrals in stages with numerical values all along the way, we don't usually pay much attention to the mechanics of the manipulations.  For an integral of this form,  
$$ \ \int \ \frac{K}{ax^2 \ + \ bx \ + \ c} \ \ dx \ \ , $$
completing the square gives us
$$ \ K \ \int \ \frac{dx}{(\sqrt{a} \ x \ + \ \frac{b}{2 \ \sqrt{a}})^2 \ + \ [ \ c \ - \ \frac{b^2}{4a} \ ]}  \ \ = \ \ K \ \int \ \frac{dx}{(\sqrt{a} \ x \ + \ \frac{b}{2 \ \sqrt{a}})^2 \ + \ \mathcal{C}}  $$
$$ = \ \ \frac{K}{\mathcal{C}} \ \int \ \frac{dx}{(\sqrt{\frac{a}{\mathcal{C}}} \ x \ + \ \frac{b}{2 \ \sqrt{a \ \mathcal{C}}})^2 \ + \ 1} \ \ , $$
and the substitution $ \ u \ = \ \sqrt{\frac{a}{\mathcal{C}}} \ x \ + \ \frac{b}{2 \ \sqrt{a \ \mathcal{C}}} \ $ gives us
$$ = \ \ \frac{K}{\mathcal{C}} \ \cdot \ \sqrt{\frac{\mathcal{C}}{a}} \ \int \ \frac{du}{u^2 \ + \ 1} \ \ = \ \ \frac{K}{\sqrt{a \ \mathcal{C}}}   \ \int \ \frac{du}{u^2 \ + \ 1} \ \ . $$
For your integral, $ \ K \ = \ 10 \ , \ a \ = \ 4 \ , $ and $ \ \mathcal{C} \ = \ 61 \ - \ \frac{24^2}{4 \cdot 4} \ =\ 25 \ $ ; the factor you called $ \ z \ $ is $ \ \frac{1}{\sqrt{a \ \mathcal{C}}} \ $ .
