<s>[INST] Rational function integration (not sure wich technique to use) I have the integral $$\int \frac{x^2 - 11x -2}{(x-2)(x+2)(2x+1)} dx $$
From what I have learned in class, it seems that I need to use polynomial long division
is that the correct attitude for that problem? and if yes when I try to devide I have problem because the deg of $x^2 - 11x -2$$\ <$ $(x-2)(x+2)(2x+1)$ but on the other hand I can't use the "Partial fraction decomposition" technique here (I probably wrong)
I would like to get some help here 
Thank you kindly
 [/INST] Basically, it can be solved by using partial fraction decomposition : 
\begin{equation}
\begin{array}{c}
=\int\left(\frac{71}{21(2 x+11)}-\frac{6}{7(x+2)}-\frac{1}{3(x-2)}\right) \mathrm{d} x \\
=\frac{71}{21} \int \frac{1}{2 x+11} \mathrm{d} x-\frac{6}{7} \int \frac{1}{x+2} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x-2} \mathrm{d} x \\
\int \frac{1}{2 x+11} \mathrm{d} x \\
\text { Substitute } u=2 x+11 \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=2 \text { (steps) } \longrightarrow \mathrm{d} x=\frac{1}{2} \mathrm{d} u \\
=\frac{1}{2} \int \frac{1}{u} \mathrm{d} u
\end{array}
\end{equation}
Plug in solved integrals:
$\frac{1}{2} \int \frac{1}{u} \mathrm{d} u$
$$
=\frac{\ln (u)}{2}
$$
Undo substitution $u=2 x+11$
$$
=\frac{\ln (2 x+11)}{2}
$$
Now solving:
$$
\begin{array}{c}
\int \frac{1}{x+2} \mathrm{d} x \\
\text { Substitute } u=x+2 \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=1_{(\text {steps })} \longrightarrow \mathrm{d} x=\mathrm{d} u:
\end{array}
$$
$=\int \frac{1}{u} \mathrm{d} u$
Use previous result:
$=\ln (u)$
Undo substitution $u=x+2$ $=\ln (x+2)$
\begin{equation}
\begin{array}{l}
\frac{71}{21} \int \frac{1}{2 x+11} \mathrm{d} x-\frac{6}{7} \int \frac{1}{x+2} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x-2} \mathrm{d} x \\
\quad=\frac{71 \ln (2 x+11)}{42}-\frac{6 \ln (x+2)}{7}-\frac{\ln (x-2)}{3}
\end{array}
\end{equation}
</s>