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?

By factoring numbers out of the denominator, we write
\begin{align*}
\int \frac{1}{4(x - 3)^2 + 25} dx &= \frac{1}{25} \int \frac{1}{\frac{4}{25} (x - 3)^2 + 1} dx \\
&= \frac{1}{25} \int \frac{1}{\left(\frac{2(x - 3)}{5}\right)^2+1} dx 
\end{align*}
Now substitute $u = 2(x - 3) / 5$, so that
$$dx = \frac 5 2 du$$
Now notice that $$\frac 1 {25} \cdot \frac 5 2 = \frac 1 {10}$$
as desired.
