See first Arturo's excellent answer to Integration by partial fractions; how and why does it work?
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.
I don't know what an indice or natural number is and it is not
mentioned naywhere in the text. (in a comment)
The numbers $i$ and $j$ are natural numbers, i.e. they are positive integers $1,2,3,\dots,n,\dots .$ Their set is denoted by $\mathbb{N}$.
$A$, $B$ 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.
In this context the leters $A$, $B$ and $C$ are constants, i.e. independent of the variable $x$.
Let
$$\begin{equation*}
\frac{P(x)}{Q(x)}:=\frac{x^{2}-5x+16}{\left( 2x+1\right) \left( x-2\right)
^{2}}\tag{1}.
\end{equation*}$$ The denominator $Q(x):=\left( 2x+1\right) \left( x-2\right) ^{2}$ has factors of the form $(ax+b)^{i}$ only. Each one originates $i\in\mathbb{N}$ partial fractions whose integrals can be computed recursively and/or found in tables of integrals. See $(6),(7),(8)$ bellow for the present case.)
$$\begin{equation*}
\frac{A_{i}}{(ax+b)^{i}}+\frac{A_{i-1}}{(ax+b)^{i-1}}+\ldots +\frac{A_{1}}{ax+b}.
\end{equation*}\tag{2}$$ The exponent of the factor $\left( x-2\right) ^{2}$ is $i=2$ and of the factor $2x+1$ is $i=1$. Therefore we should find the constants $A_{1}$, $A_{2}$, $B$ such that
$$\begin{equation*}
\frac{P(x)}{Q(x)}=\frac{x^{2}-5x+16}{\left( 2x+1\right) \left( x-2\right)
^{2}}=\frac{B}{2x+1}+\frac{A_{2}}{\left( x-2\right) ^{2}}+\frac{A_{1}}{x-2}\end{equation*}.\tag{3}$$
One method† is to reduce the RHS to a common denominator
$$\begin{equation*}
\frac{x^{2}-5x+16}{\left( 2x+1\right) \left( x-2\right) ^{2}}=\frac{B\left(x-2\right) ^{2}+A_{2}\left( 2x+1\right) +A_{1}\left( x-2\right) \left(2x+1\right) }{\left( 2x+1\right) \left( x-2\right) ^{2}}.
\end{equation*}$$ $$\tag{3a}$$
[See remak below.] This means that the polynomials of the numerators must be equal on both sides of this last equation. Expanding the RHS, grouping the terms of the same degree
$$\begin{eqnarray*}
P(x) &:=&x^{2}-5x+16=B\left( x-2\right) ^{2}+A_{2}\left( 2x+1\right)
+A_{1}\left( x-2\right) \left( 2x+1\right) \\
&=&\left( Bx^{2}-4Bx+4B\right) +\left( 2A_{2}x+A_{2}\right) +\left(
2A_{1}x^{2}-3A_{1}x-2A_{1}\right) \\
&=&\left( B+2A_{1}\right) x^{2}+\left( -4B+2A_{2}-3A_{1}\right) x+\left(
4B+A_{2}-2A_{1}\right)
\end{eqnarray*}$$ $$\tag{3b}$$ and equating the coefficients of $x^{2}$, $x^{1}$ and $x^{0}$, we conclude that they must satisfy‡ the following system of 3 linear equations [See (*) for a detailed solution of the system]
$$\begin{equation*}
\left\{
\begin{array}{c}
B+2A_{1}=1 \\
-4B+2A_{2}-3A_{1}=-5 \\
4B+A_{2}-2A_{1}=16
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
A_{1}=-1 \\
A_{2}=2 \\
B=3.
\end{array}
\right.\tag{3c}
\end{equation*}$$ In short, this method reduces to solving a linear system. So, we have
$$\begin{equation*}
\frac{x^{2}-5x+16}{\left( 2x+1\right) \left( x-2\right) ^{2}}=\frac{3}{2x+1}+
\frac{2}{\left( x-2\right) ^{2}}-\frac{1}{x-2}.
\end{equation*}\tag{4}$$
We are now left with the integration of each partial fraction
$$\begin{equation*}
\int \frac{x^{2}-5x+16}{\left( 2x+1\right) \left( x-2\right) ^{2}}dx=3\int
\frac{1}{2x+1}dx+2\int \frac{1}{\left( x-2\right) ^{2}}dx\\-\int \frac{1}{x-2}
dx.\tag{5}
\end{equation*}$$
Can you proceed from here? Remember these basic indefinite integral formulas:
$$\int \frac{1}{ax+b}dx=\frac{1}{a}\ln \left\vert ax+b\right\vert +C, \tag{6}$$
$$\int \frac{1}{\left( x-r\right) ^{2}}dx=-\frac{1}{x-r}+C,\tag{7}$$
$$\int \frac{1}{x-r}dx=\ln \left\vert x-r\right\vert +C.\tag{8}$$
--
† Another method is to evaluate both sides of $(3)$ at 3 different values, e.g. $x=-1,0,1$ and obtain a system of 3 equations. Another one is to compute $P(x)$
$$\begin{equation*}
P(x)=x^{2}-5x+16=B\left( x-2\right) ^{2}+A_{2}\left( 2x+1\right)
+A_{1}\left( x-2\right) \left( 2x+1\right)
\end{equation*}$$
first at the zeros of each term, i.e. $x=2$ and $x=-1/2$
$$\begin{eqnarray*}
P(2) &=&10=5A_{2}\Rightarrow A_{2}=2 \\
P\left( -1/2\right) &=&\frac{75}{4}=\frac{25}{4}B\Rightarrow B=3;
\end{eqnarray*}$$
and then at e.g. $x=0$
$$\begin{equation*}
P(0)=16=4B+A_{2}-2A_{1}=12+2-2A_{1}\Rightarrow A_{1}=-1.
\end{equation*}$$
For additional methods see this Wikipedia entry
‡ If $B+2A_{1}=1,-4B+2A_{2}-3A_{1}=-5,4B+A_{2}-2A_{1}=16$, then $x^{2}-5x+16=\left( B+2A_{1}\right) x^{2}+\left( -4B+2A_{2}-3A_{1}\right) x+\left(4B+A_{2}-2A_{1}\right)$ for all $x$ and $(3a)$ is an identity.
REMARK in response a comment below by OP. For $x=2$ the RHS of $(3a)$ is not defined. But we can compute as per $(3b,c)$ or as per †, because we are not plugging $x=2$ in the fraction $(3a)$. In $(3c)$ we assure that the numerators of $(3a)$ $$x^{2}-5x+16$$ and $$B\left( x-2\right) ^{2}+A_{2}\left( 2x+1\right)
+A_{1}\left( x-2\right) \left( 2x+1\right) $$ are identically equal, i.e. they must have equal coefficients of $x^2,x,x^0$.
(*) Detailed solution of $(3c)$. Please note that we cannot find $A,B$ and $C$ with one equation only, as you tried below in a comment ("$16=2b+A_1−A_2$ I have no idea how to solve this.")
$$\begin{eqnarray*}
&&\left\{
\begin{array}{c}
B+2A_{1}=1 \\
-4B+2A_{2}-3A_{1}=-5 \\
4B+A_{2}-2A_{1}=16
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
B=1-2A_{1} \\
-4\left( 1-2A_{1}\right) +2A_{2}-3A_{1}=-5 \\
4\left( 1-2A_{1}\right) +A_{2}-2A_{1}=16
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
B=1-2A_{1} \\
-4+5A_{1}+2A_{2}=-5 \\
4-10A_{1}+A_{2}=16
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
B=1-2A_{1} \\
A_{2}=-\frac{1+5A_{1}}{2} \\
4-10A_{1}-\frac{1+5A_{1}}{2}=16
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
B=1-2A_{1} \\
A_{2}=-\frac{1+5A_{1}}{2} \\
A_{1}=-1
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
B=1-2\left( -1\right) \\
A_{2}=-\frac{1+5\left( -1\right) }{2} \\
A_{1}=-1
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
A_{1}=-1 \\
A_{2}=2 \\
B=3
\end{array}
\right.
\end{eqnarray*}$$
Comment below by OP
I watched the MIT lecture on this and they use the "cover up" method to solve systems like this and I am attempting to use that here. I have $$\frac{A}{2x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$$ Is there anything wrong so far? It appears to me to be correct. Now I try to find B by making x = 2 and multplying by x-2 which gets rid of C and A since it makes them zero and then the RHS which cancels out and leaves me with B = 2 but that also works for C I think so I am confused, and for A I get 55/6 which I know is wrong but the method works and I am doing the math right so what is wrong?
Starting with $$\frac{x^{2}-5x+16}{(2x+1)(x-2)^{2}}=\frac{A}{2x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}}\tag{3'}$$
we can multiply it by $(x-2)^{2}$
$$\frac{x^{2}-5x+16}{2x+1}=\frac{A(x-2)^{2}}{2x+1}+B(x-2)+C.$$
To get rid of $A$ and $B$ we make $x=2$ and obtain $C$
$$\frac{2^{2}-5\cdot 2+16}{2\cdot 2+1}=\frac{A(2-2)^{2}}{2x+1}+B(2-2)+C$$
$$\Rightarrow 2=0+0+C\Rightarrow C=2$$
We proceed by multiplying $(3')$ by $2x+1$
$$\frac{x^{2}-5x+16}{(x-2)^{2}}=A+\frac{B(2x+1)}{x-2}+\frac{C(2x+1)}{(x-2)^{2}}$$
and making $x=-1/2$ to get rid of $B$ and $C$
$$\frac{\left( -1/2\right) ^{2}-5\left( -1/2\right) +16}{(-1/2-2)^{2}}=A+
\frac{B(2\left( -1/2\right) +1)}{-1/2-2}+\frac{C(2\left( -1/2\right) +1)}{
(-1/2-2)^{2}}$$
$$\Rightarrow 3=A+0+0\Rightarrow A=3$$
Substituing $A=3,C=2$ in $(3')$, we have
$$\frac{x^{2}-5x+16}{(2x+1)(x-2)^{2}}=\frac{3}{2x+1}+\frac{B}{x-2}+\frac{2}{
(x-2)^{2}}$$
Making e.g. $x=1$ (it could be e.g. $x=0$)
$$\frac{1^{2}-5+16}{(2+1)(1-2)^{2}}=\frac{3}{2+1}+\frac{B}{1-2}+\frac{2}{
(1-2)^{2}},$$
$$\Rightarrow 4=1-B+2\Rightarrow B=-1.$$
Thus
$$\frac{x^{2}-5x+16}{(2x+1)(x-2)^{2}}=\frac{3}{2x+1}-\frac{1}{x-2}+\frac{2}{(x-2)^{2}},\tag{3''}$$
which is the same decomposition as $(4)$.
$$environment in the title. – Asaf Karagila May 31 '12 at 11:53