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$$\ <$ $(x-2)(x+2)(2x+1)$ but on the other hand I can't use the "Partial fraction decomposition" technique here (I probably wrong)
I would like to get some help here 
Thank you kindly
 A: The degree of the numerator is less than the degree of the denominar, so you don't need to do any poynomial long division. You need to decompose the integrand into partial fractions, $i.e.$ you need to find real numbers $a,b,c$ such that $$\frac{x^2 - 11x -2}{(x-2)(x+2)(2x+1)}=\frac{a}{x-2}+\frac{b}{x+2}+\frac{c}{2x+1}.$$
A: Basically, it can be solved by using partial fraction decomposition : 
\begin{equation}
\begin{array}{c}
=\int\left(\frac{71}{21(2 x+11)}-\frac{6}{7(x+2)}-\frac{1}{3(x-2)}\right) \mathrm{d} x \\
=\frac{71}{21} \int \frac{1}{2 x+11} \mathrm{d} x-\frac{6}{7} \int \frac{1}{x+2} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x-2} \mathrm{d} x \\
\int \frac{1}{2 x+11} \mathrm{d} x \\
\text { Substitute } u=2 x+11 \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=2 \text { (steps) } \longrightarrow \mathrm{d} x=\frac{1}{2} \mathrm{d} u \\
=\frac{1}{2} \int \frac{1}{u} \mathrm{d} u
\end{array}
\end{equation}
Plug in solved integrals:
$\frac{1}{2} \int \frac{1}{u} \mathrm{d} u$
$$
=\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}
