<s>[INST] How is this is an identity? So we have started studying partial fractions.The book teaches two methods: 


*

*By equating coefficients

*By utilizing the fact that when a rational fraction is decomposed to partial fractions it is an identity
I cannot quite understand how does the 2nd method works.If we make the denominator of one of the fractions zero , it would become undefined , so how will it be an identity then???
Thanks in Advance.

For example, take the following
$$
\frac{1}{(x+3)(x+4)}=\frac{a}{(x+3)} + \frac{b}{(x+4)}
$$
so if we put $x=-3$ then the left hand side of the equation would become undefined.
Also on the right hand side the term $\frac{a}{(x+3)}$ would also become undefined, so how would this equation be an identity?
 [/INST] Partial Fractions : Steps
Step (1) Check the degrees of both numerator (n) and denominator (d) separately.
(i)  If $n<d$ follow step (2)
(ii) If $n \geq d$ Add a polynomial of degree $(n-d)$, and follow (2)
Ex : $\frac{x^2}{x^3+1} \Rightarrow 2<3 \Rightarrow $ follow step (2)
Ex: $\frac{x^3+x+1}{x^3-2x-1} \Rightarrow 3=3 \Rightarrow \frac{x^3+x+1}{x^3-2x-1} =A+ $ .....follow step (2)
Ex: $\frac{3x^4+x^2+1}{3x^3-2x+1} \Rightarrow 4>3\Rightarrow \frac{3x^4+x^2+1}{3x^3-2x+1} =Ax+B+ $ ......follow step (2)
Ex: $\frac{x^5+x+1}{3x^3-2x+1} \Rightarrow 5>3\Rightarrow \frac{x^5+x+1}{3x^3-2x+1} =Ax^2+Bx+C +$ .......follow step (2)
Step (2) : Factorize the denominator and write each factor separately with a numerator of degree less than one with respect to the chosen factor.
Ex: $\frac{x^5+x+1}{3x^3-2x+1} \Rightarrow 5>3\Rightarrow \frac{x^5+x+1}{3x^3-2x+1} =\frac{x^5+x+1}{(x+1)(3x^2-3x+1)} =Ax^2+Bx+C + \frac {D}{(x+1)}+ \frac{Ex+F}{(3x^2-3x+1)}$  
Step (3) : Now find $A,B,C,D,E,F$ by multiplying whole expression by $(x+1)(3x^2-3x+1)$ . You could use $x=-1$ to find constants and compare co-efficients.
Important Ex: 
$1) \frac{x}{(x+1)^3} = \frac {A}{(x+1)}+\frac{B}{(x+1)^2}+\frac {C}{(x+1)^3}$
$2) \frac{x}{(x^2+1)^3} = \frac {Ax+B}{(x^2+1)}+\frac{Cx+D}{(x^2+1)^2}+\frac {Ex+F}{(x^2+1)^3}$
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