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In order to integrate the following function, our professor taught us the following method:

$\dfrac{x^4+1}{x^3+x^2} = \dfrac{x^4+1}{x^2(x+1)}=\dfrac{A}{x^2}+\dfrac{B}{x}+\dfrac{C}{x+1}$

I don't understand why we can do that, in reality, $\dfrac{x^4+1}{x^3+x^2} = \dfrac{x^4+1}{xx(x+1)}=\dfrac{A}{x}+\dfrac{B}{x}+\dfrac{C}{x+1}$

Although I understand why we cannot find $A, B$and$C$ that way (Indeed to find A we evaluate $\dfrac{x^4+1}{x^2(x+1)}$ at x=0)

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    $\begingroup$ $\frac{A}{x}$ and $\frac{B}{x}$ are not linearly independent. $\frac{A}{x}+\frac{B}{x}=\frac{A+B}{x}=\frac{k}{x}$. $\endgroup$ – GoodDeeds May 3 '16 at 16:45
  • $\begingroup$ This method (simple fractions) is based on the fact, that is usually studied in a later course in abstract algebra (Fields extensions and/or Galois theory) that any real polynomial can always be written as a product of linear and/or quadratic polynomials. That's why this thing works. As for why the way of writing the denominators, GoodDeeds already explained you. $\endgroup$ – DonAntonio May 3 '16 at 16:48
  • $\begingroup$ This questions seems to be less about integration and more about the intuition (or perhaps a proof of) partial fraction decomposition. $\endgroup$ – zahbaz May 3 '16 at 16:50
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    $\begingroup$ Since the degree of the polynomial in the numerator is greater than that of the divisor, I believe the method requires to first conduct polynomial division amongst the two and then proceed in the manner described. $\endgroup$ – MathematicianByMistake May 3 '16 at 16:50
  • $\begingroup$ There is an extremely detailed answer here that may help you out: math.stackexchange.com/questions/20963/… $\endgroup$ – wgrenard May 3 '16 at 16:51
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You can see that your partial fraction decomposition cannot work simply noting that the numerator of the fraction: $A(x+1)+Bx(x+1)+Cx^2$ cannot be a polynomial of degree $4$. In other words, if the denominators are the factors of a polynomial of degree $n$, and the numerators of the partial fractions are numbers, that the numerator of their sum cannot be of degree $\ge n$.

As noted in the comments, the first step is to use long division to find: $$ \frac{x^4+1}{x^3+x^2}=x-1+\frac{x^2+1}{x^3+x^2} $$ than we can use partial fractions for: $$ \frac{x^2+1}{x^2(x+1)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C}{x+1} $$

and find $A=1,B=-1,C=2$

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