# Partial fraction: is my final answer and computation correct?

I need some help at this exercise (partial fraction): $$\int{\frac{4x + 1}{x^3(x+2)}} dx$$

First of all I calculated the roots of the denominator. $$x_{1, 2, 3} = 0 \\ x_4 = -2$$

After that I assigned the roots to the partial fraction. $$\frac{A_1}{x}, \frac{A_2}{x^2}, \frac{A_3}{x^3}, \frac{B}{(x+2)} \\ \frac{4x + 1}{x^3(x+2)} = \frac{4x + 1}{x^3(x+2)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x^3} + \frac{B}{(x+2)} \\ 4x + 1 = A_1x^2(x+2) + A_2x(x+2) + A_3(x+2) + Bx^3 \\ 4x + 1 = A_1(x^3+2x^2) + A_2(x^2+2x) + A_3(x+2) + Bx^3 \\$$

Then I insert the root values. For $-2$, I get $B = \frac{7}{8}$. For $0$ I get $A_3 = \frac{1}{2}$. After that I insert two additional numbers, I choose $1$ and $-1$. I transformend the equation to $A_1$ and $A_2$ and get $A_1 = \frac{23}{20}$ and $A_2 = \frac{197}{120}$.

$$\frac{4x + 1}{x^3(x+2)} = \frac{\frac{23}{20}}{x} + \frac{\frac{197}{120}}{x^2} + \frac{\frac{1}{2}}{x^3} + \frac{\frac{7}{8}}{(x+2)}$$

Could anyone tell me if it's correct? I belive it is wrong :), where is my misstake?

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 It should be $\frac{A_3}{x^3}$ not $\frac{A_3}{x^2}$ – Thomas Andrews Aug 27 '12 at 11:45 Oh yea. I mistype. Sorry. – hofmeister Aug 27 '12 at 11:49

Reduce the RHS to a common denominator

$$\begin{eqnarray*} \frac{4x+1}{x^{3}(x+2)} &=&\frac{A_{1}}{x}+\frac{A_{2}}{x^{2}}+\frac{A_{3}}{ x^{3}}+\frac{B}{x+2} \\ &=&\frac{A_{1}x^{2}\left( x+2\right) +A_{2}x\left( x+2\right) +A_{3}\left( x+2\right) +Bx^{3}}{x^{3}(x+2)} \\ &=&\frac{\left( A_{1}+B\right) x^{3}+\left( 2A_{1}+A_{2}\right) x^{2}+\left( 2A_{2}+A_{3}\right) x+2A_{3}}{x^{3}(x+2)}. \end{eqnarray*}$$

You can proceed by equating coefficients of $$\begin{equation*} 4x+1=\left( A_{1}+B\right) x^{3}+\left( 2A_{1}+A_{2}\right) x^{2}+\left( 2A_{2}+A_{3}\right) x+2A_{3}. \end{equation*}$$

So$^1$ $$\begin{eqnarray*} 2A_{3} &=&1\Leftrightarrow A_{3}=\frac{1}{2} \\ 2A_{2}+\frac{1}{2} &=&4\Leftrightarrow A_{2}=\frac{7}{4} \\ 2A_{1}+\frac{7}{4} &=&0\Leftrightarrow A_{1}=-\frac{7}{8} \\ -\frac{7}{8}+B &=&0\Leftrightarrow B=\frac{7}{8}, \end{eqnarray*}$$

and $$\begin{equation*} \frac{4x+1}{x^{3}(x+2)}=-\frac{7}{8}\frac{1}{x}+\frac{7}{4}\frac{1}{x^{2}}+ \frac{1}{2}\frac{1}{x^{3}}+\frac{7}{8}\frac{1}{x+2}. \end{equation*}$$

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$^1$ It corresponds to solve the system

$$\left\{ \begin{array}{c} A_{1}+B=0 \\ 2A_{1}+A_{2}=0 \\ 2A_{2}+A_{3}=4 \\ 2A_{3}=1 \end{array} \right.$$

Then I insert the root values. For $-2$, I get $B = \frac{7}{8}$. For $0$ I get $A_3 = \frac{1}{2}$. After that I insert two additional numbers, I choose $1$ and $-1$.

Using $$\begin{equation*} 4x+1=A_{1}x^{2}\left( x+2\right) +A_{2}x\left( x+2\right) +A_{3}\left( x+2\right) +Bx^{3} \end{equation*}$$

you get for $x=1$ $$\begin{equation*} 5=3A_{1}+3A_{2}+\frac{3}{2}+\frac{7}{8}, \end{equation*}$$

and for $x=-1$ $$\begin{equation*} -3=A_{1}-A_{2}+\frac{1}{2}-\frac{7}{8}, \end{equation*}$$

whose solutions are $A_{2}=\frac{7}{4},A_{1}=-\frac{7}{8}$ as above and not $A_1 = \frac{23}{20}$, $A_2 = \frac{197}{120}$.

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How do you get $A_2$? Also insert two additional numbers? – hofmeister Aug 27 '12 at 11:50
@hofmeister: This answer is a final complete answer. In some cases, you can find the unknown coffecients by puting the roots of the denominator into the numerator. For example, putting $x=0$ into numerator, you get $A_3(0+2)=4*0+1\longrightarrow A_3=1/2$. – Babak S. Aug 27 '12 at 11:54
@hofmeister $2A_{2}+A_{3}=4\Leftrightarrow 2A_{2}+\frac{1}{2}=4\Leftrightarrow 2A_{2}=4-% \frac{1}{2}\Leftrightarrow A_{2}=\frac{7}{4}$ – Américo Tavares Aug 27 '12 at 11:55
Ahhh, now i get it! Thanks. I didn't know about that. Is it recommanded to factorise $x$ everytime in such a case? Does it also work with insert values? Just to know...? – hofmeister Aug 27 '12 at 12:19
Yes I see! Thanks, I have calculate it a second time and now I get the same result. Thanks for your greate help. Especially for the tricks. – hofmeister Aug 27 '12 at 13:57