<s>[INST] 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?
 [/INST] Given
$$
\int \frac{10}{4 x^2 - 24 x + 61} dx
$$
Rewrite as
$$
\int \frac{10}{4 x^2 - 24 x + 61} dx =
\int \frac{10}{ \big( 2x - 6 + 5 \textbf{i} \big) \big( 2x - 6 - 5 \textbf{i} \big) } dx
$$
So
$$
\int \frac{10}{4 x^2 - 24 x + 61} dx =
\textbf{i} \int \left(
\frac{1}{ 2x - 6 + 5 \textbf{i} } - \frac{1}{  2x - 6 - 5 \textbf{i}  }
\right) dx
$$
Thus
$$
\int \frac{10}{4 x^2 - 24 x + 61} dx =
- \textbf{i} \ln \left(
\frac{ \displaystyle 1 + \frac{2x-6}{5} \textbf{i} }{ \displaystyle 1 - \frac{2x-6}{5}\textbf{i} }
\right) = -\textbf{i} \tanh^{-1} \left( \frac{2x-6}{5} \textbf{i} \right) = \tan^{-1}\left( \frac{2x-6}{5} \right)
$$

$$
\int_3^{11/2} \frac{10}{4 x^2 - 24 x + 61} dx = \left[ \tan^{-1}\left( \frac{2x-6}{5} \right)\right]_3^{11/2} = \tan^{-1}\Big(1\Big) = \frac{\pi}{4}
$$
</s>