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Where is the mistake in calculations for $\int \frac {2x+1}{(x-3)^2(x+1)}$ In here, theoretically you confirm me that I can use this partial fraction decomposition of fraction and I wanted to apply for that integral.

First way:
$\frac {2x+1}{(x-3)^2(x+1))}=\frac{A}{x+1} + \frac{B}{(x-3)} + \frac{C}{(x-3)^2} $ and at the final I found coefficients as $A=-1/16,B=1/16 , C= 7/4$ (Which is correct as I checked in calculator.)

Second Way: $\frac {2x+1}{(x-3)^2(x+1))}=\frac {A}{x+1} + \frac {Bx+C}{(x-3)^2}$ (because I thought since I divide something by degree $2$ which is $(x-3)^2$ so remainder at most can be degree $1$) and at the final I couldn't find same coefficients as above. Again $A$ is $-1/16$ and $B=1/16$ but I have two results for $C$ so that leads to contradiction.

I got $A,B,C$ from: $A(x-3)^2 + (Bx+C)(x+1)=Ax^2-6Ax+9A+Bx^2+Bx+Cx+C=2x+1$

Where is the mistake? What am I missing?

Fuat Ray
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    What do you mean "two values for $C$"? Your "second way" just (mistakenly) combines $\frac{B}{x-3}+\frac{C}{(x-3)^2}$ into a single fraction; so you will get $$\frac{bx+c}{(x-3)^2} = \frac{B}{x-3}+\frac{C}{(x-3)^2} = \frac{B(x-3)+C}{(x-3)^2},$$ which tells you that you will get $b=B$ and $c=C-3B$. There is no "contradiction". – Arturo Magidin May 25 '23 at 18:42
  • I mean I managed to find two results for $C$. From $9A+C=1$ and other from $-6A+B+C=2$. @ArturoMagidin Meanwhile, is the idea of second way true? – Fuat Ray May 25 '23 at 18:52
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    You can write it that way (again, you are just expressing $B/(x-3)+C/(x-3)^2$ as a single fraction), but you will have to break it up again to do the integral, so it is wasted work. But I don't see why you say you get "two values" for $C$. You have $A+B=0$, $B+C-6A=2$, and $9A+C=1$. This has a unique solution for $A$, $B$, and $C$. You have $A=-B$, and then $C=2+7A$, $C=1-9A$. This gives $2+7A=1-9A$, or $16A=-1$, so $A=-1/16$; $B=1/16$; and then either equation gives the same value of $C$: $2+7(-1/16) = 2-(7/16) = 25/16$, $1-9A=1+(9/16)=25/16$. Equal. – Arturo Magidin May 25 '23 at 19:05
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    Now, the problem is that you are using the same notation for two different expressions. Note that the values of $A$, $B$, and $C$ when you express it as $\frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$ are not the same as the values of $a,b,c$ when you express it as $\frac{a}{x+1} + \frac{bx+c}{(x-3)^2}$, because, as I said in my first comment, these are different expressions. While you will get $a=A$ and $b=B$ in this particular instance, you will get $c=C-3B$; that is, $c\neq C$. But you don't have two values for the same constant, you have two different constants. – Arturo Magidin May 25 '23 at 19:10
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    Perhaps because you insist on calling both of them $C$ you think they should be equal? They aren't. They are the result of different expressions, so there is no reason why they should be the same number. – Arturo Magidin May 25 '23 at 19:12
  • Thank you very much. I made computation mistake and understand totally. Also, actually they should be same up to constant. I am glad to confirm that I can use the idea that I mention – Fuat Ray May 25 '23 at 19:15
  • You can "use the idea", but it is a waste of time. How are you going integrate $\frac{bx+c}{(x-3)^2}$, if not by rewriting it as $\frac{B}{x-3} + \frac{C}{(x-3)^2}$? There's a reason that the algorithm is to write it in the latter way from the start. – Arturo Magidin May 25 '23 at 19:22
  • It doesn't seems to me that natural maybe because I don't understand the idea of it. As I said ago, my idea is coming from division and I can't break the ice for why we keep continuing to write term like you written $\frac{B}{x-3} + \frac{C}{(x-3)^2}$. Basically, if we have something of power $5$ we need to write up to $\frac {F}{(x-3)^5}$ – Fuat Ray May 25 '23 at 19:27
  • Because the point is to be able to do the integral. You cannot do the integral $\int\frac{bx+c}{(x-3)^2},dx$ with a simple substitution; instead you end up having to break it up and doing three integrals, after doing a substituion $u=x-3$ in $\int\frac{bx}{(x-3)^2},dx$ and in $\int\frac{c}{(x-3)^2},dx$. So in fact you are really doing an integral of the form $\int\frac{B}{x-3},dx$ and another of the form $\int\frac{C}{(x-3)^2},dx$. So by doing it directly that way, you save a lot of unnecessary work. That's the idea: to actually integrate. – Arturo Magidin May 25 '23 at 19:32
  • So, what I understand from your writings is we want to write expressions in terms of expression that we can take integral of it easily. For example, if we have $\frac{C}{(x-3)^2}$ then idea is how can I write this in terms of sum of expression that I can take integral. Try to write it as sum of $\frac{A}{(x-3)}$ ( we know its integration) and $\frac{B}{(x-3)^2}$ (also we know its integraiton)and there is no need to continue as we have degree $2$. – Fuat Ray May 25 '23 at 19:38
  • You need to review the point of partial fraction decomposition. You are evidently lost. – Arturo Magidin May 25 '23 at 19:40
  • You may be right. So, you're saying what I wrote is wrong in a minute ago – Fuat Ray May 25 '23 at 19:44
  • I'm saying that you are clearly not understanding what is the point of partial fraction decomposition, given that you don't "understand the idea of it". It is useless to spend a long cramped comment thread with the end results, given that you do not understand the point to begin with. – Arturo Magidin May 25 '23 at 19:46

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