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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.

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Please avoid using $$ environment in the title. –  Asaf Karagila May 31 '12 at 11:53
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Just a friendly advice: If you really think that i and j in that equation do stand for some imaginary numbers, you should probly invest more time in understanding the context of your exercise. As those are usually just indices and therefore natural numbers. :-) –  user32610 May 31 '12 at 11:59
    
Have you looked at this discussion yet? –  Arturo Magidin May 31 '12 at 21:00
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"$A$, $B$ 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." $A$, $B$ and $C$ are constants, thus independent of the variable $x$. –  Américo Tavares Jun 6 '12 at 16:21
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Americo's comment is helpful and informative, not offensive, so whoever flagged it as such is mistaken. –  Zev Chonoles Jun 7 '12 at 0:51
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3 Answers

up vote 8 down vote accepted

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 methodis 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 $(4)$.

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I really can not follow what you are trying to say in this. What does " Each one originates i∈N partial fractions which are easy to integrate" mean? –  user138246 Jun 4 '12 at 20:02
    
I do not understand the part where you actually split the problem into 3 fractions, I get the part with 2x+ 1 and (x-2)^2 but why is there then another x-2 denominator? –  user138246 Jun 4 '12 at 20:36
    
Because there is an algebraic theorem that states that the expansion into partial fractions has (in the denominator) not only the term with higher exponent $(x-2)^2$ but also the linear term $(x-1)$(exponent $1$). (For the exponent $i$ the same theorem requires $i$ partial fractions.) –  Américo Tavares Jun 4 '12 at 20:41
    
So this is just something I should memorize and not worry about understanding? –  user138246 Jun 4 '12 at 20:43
    
You might understand it but the proof is not that easy. (In a book I have it has 3 pages). –  Américo Tavares Jun 4 '12 at 20:46
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Rather than continue to give explicit hints on your homework problems for this assignment, I am going to treat this question as a for partial fractions.

My preferred site reference for partial fractions is Paul's Online Math Notes. It's a great thing to know about and consider when you're learning calculus.

It has good exposition, lots of examples, explicit if-then problem solving plans, and is overall a great reference.

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Also, just remember that the idea behind partial fractions is to 'undo' combining fractions by finding a common denominator, etc. --- essentially, you're trying to take one big fraction, which is hard to integrate, and replace it by the sum of simpler fractions, which are easier to integrate. –  John Engbers May 31 '12 at 11:41
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+1 Very nice and educative answer. No need to solve for him or even hints. That Paul's site is very clear. –  DonAntonio May 31 '12 at 11:56
    
I was just trying to get a quick run down of the problem before class because this is a homework question and I need full points on the homework because I will fail several tests in this class. –  user138246 May 31 '12 at 11:57
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"...because you will fail several tests in this class"? Sir, your optimism is overwhelming. –  DonAntonio May 31 '12 at 13:39
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Math classes are very difficult for people who aren't naturally good at math. It takes more time to study and even then tests are very stressful and I always forget things I know on a test and make mistakes, no matter what. So I need that buffer of other points because I will get a D or C average on the tests. –  user138246 Jun 1 '12 at 13:08
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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.

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Aren't you mising an x-2 term? –  user138246 Jun 4 '12 at 20:46
    
What is Bx and why isn't there an Ax? –  user138246 Jun 4 '12 at 22:48
    
I do not follow from \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*} –  user138246 Jun 4 '12 at 22:49
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