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How to find out the partial fraction decomposition form (or pattern ) of a rational function $\dfrac{P(X)}{Q(x)}$ , for example we know that the partial fraction decomposition form of $$\frac{x^{2}+3x+1}{(x-1)^{2}(x+3)},$$ would be something like this : $$\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+3}, \qquad A,B,C \in \mathbb{R}$$

but how did we "guess" that pattern ? and how to guess it in general case ?

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    You have to factor the denominator. It is not an easy problem in general, but in exercises, you often want to find rational roots of the denominator, so you can use the Rational Root Test on $Q(x)$. – 2017-01-12
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    It seems to me you know how to find the partial fraction decomposition, but your question is more about why the decomposition is the way it is. You might want to clarify that with a different title. – 2017-01-12
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    If you don't receive a satisfying answer in the next couple hours, ping me (address me in a comment, including @amWhy), and I'll work on this with you, okay, Mamadou? – 2017-01-12
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    Are you still needing help on this, @Mamadou ? – 2017-01-14

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We are just doing polynomial long division on the numberator, and collecting the remainders.

$\frac {p(x)}{(x+3)(x-1)^2} = \frac {q(x) (x+3) + r}{(x+3)(x-1)^2} = \frac {r}{(x+3)} + \frac {q(x)+r}{(x-1)^2}$

$r$ will always have degree less than its denominator. $q(x)$ will have degree less than $p(x).$

Next we do polynomial division to $q(x)+r$ breaking into $\frac {r_2(x-1) + r_3}{(x-1)^2} = \frac {r_2}{x-1} + \frac {r_3}{(x-1)^2}$

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    No problem. :-) See my comment immediately after your comment. – 2017-01-12
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    I need a proof-reader. – 2017-01-12
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    hahahah... I hear you; don't we all! – 2017-01-12
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First start off with the following:

$\frac{x^2+3x+1}{(x-1)^2(x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}$

Clear the denominator

$x^2+3x+1=A(x-1)(x+3)+B(x+3)+C(x-1)^2$

Expand and Collect:

$x^2+3x+1=A(x^2+2x-3)+B(x+3)+C(x^2-2x+1)$
$x^2+3x+1=(A+C)x^2+(2A+B-2C)x+(3B+C-3A)$

Use the following Equations:
$A+C=1$
$2A+B-2C=3$
$3B+C-3A=1$

Add up the equations to get $B=5/4$. This one is going to have fractions.

If anything is unclear, I can go into more depth.

For details on picking what would go in the numerator, like when to use an $Ax+B$, go here.

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    The OP seems to be asking how and why you conclude what you took for granted in your first step. – 2017-01-12
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    I know how to find the A , B and C terms the problem is how to guess that the decomposition was in this form :$~~\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}~~ ???$ suppose that we had $ \frac{x^{3}}{(x^{2}+3)^{2}(x+5)(x-4)} $ what would be the decomposition form like ? is it : $~~\frac{A}{x+3}+\frac{Bx+D}{(x+3)^2}+\frac{C}{x+5}\frac{G}{x-4}~~ $ $ ? how to know in which manner to decompose it ??? – 2017-01-12
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    @amWhy that's it ! it's exactly whats I was speaking about . – 2017-01-12
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    That's exactly what it will look like. It will also have nasty looking coefficients. – 2017-01-12
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    @Mamadou I have explained the mathematical reasoning for the decomposition in my answer. – 2017-01-12
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    @emka The OP is asking **how and why** you chose the partial fractions given the original quotient. – 2017-01-12
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    You have not explained the mathematical reasoning for how you determined how to break down $\frac{x^2+3x+1}{(x-1)^2(x+3)}$ to get $ \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}$. If I understand the OP correctly, **this is precisely what they are asking about.** – 2017-01-12
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Resolve the denominator into factors, break them,and in numerators of each write an arbitrary polynomial expression with degree one less than the denominator. Find it by comparison.