Integral of $\int \frac{x^2 - 5x + 16}{(2x+1)(x-2)^2}dx$ I am trying to find the integral of this by using integration of rational functions by partial fractions.
$$\int \frac{x^2 - 5x + 16}{(2x+1)(x-2)^2}dx$$
I am not really sure how to start this but the books gives some weird formula to memorize with no explanation of why $\frac {A}{(ax+b)^i}$ and $ \frac {Ax + B}{(ax^2 + bx +c)^j}$ 
I am not sure at all what this means and there is really no explanation of any of it, I am guessing $i$ is for imaginary number, and $j$ is just a representation of another imaginary number that is no the same as $i$. $A$, $B$ and $C$, I have no idea what that means and I am not familiar with capital letters outside of triangle notation so I am guessing that they are angles of lines for something. 

Write $$\frac{1}{(2x-1)\cdot (x-2)^{2}} = \frac{A}{2x+1} + \frac{Bx + C}{(x-2)^{2}}$$
Once you have written this down it makes the job more easier. Now, the denominator terms cancel and you are left with
\begin{align*}
1 &=  A \cdot \bigl(x-2)^{2} + B\cdot x \cdot (2x+1) + C \cdot (2x+1) \\\ 1 &= A \cdot \bigl(x^{2} - 4x +4) + 2Bx^{2} + Bx + 2Cx + C \\\ 1 &= x^{2} \cdot (A + 2B) + x   \cdot (-4A + B + 2C) +  (4A + C)
\end{align*}
Now comparing both the sides you find that 
\begin{align*}
A+2B &=0 \\\ -4A+B+2C &=0 \\\ 4A+C &=1
\end{align*}
From here find the value of $A,B$ and $C$ and try to solve the problem.
