Question :Write $$\frac{x^5}{(x^2+1 )(x+1)^2}$$ as a sum of partial fraction

What I've tried is to do polynomial long division twice to reduce the degree of numerator to be smaller than denominator than carry on to the normal steps of partial fraction decomposition

this is what i get $$x - 2 + \frac{1}{2x^2+2} + \frac{2}{x+1} - \frac{1}{2(x+1)^2}$$

but i'm not sure if its correct

  • 2
    $\begingroup$ You can verify your result simply performing the sum. $\endgroup$ – Emilio Novati Jun 2 '15 at 19:30
  • $\begingroup$ Thanks i forgot that i could check that way $\endgroup$ – Samuel Chua Jun 2 '15 at 19:34

$$x-2+\frac{1}{2(x^2+1)}+\frac{2}{x+1}-\frac{1}{2(x+1)^2}$$ $$=\frac{2(x-2)(x^2+1)(x+1)^2+(x+1)^2+4(x^2+1)(x+1)-(x^2+1)}{2(x^2+1)(x+1)^2}$$ $$=\frac{2x^5}{2(x^2+1)(x+1)^2}=\frac{x^5}{(x^2+1)(x+1)^2}$$ So, yours is correct.


Hint: $x^5 = x^3(x^2+1) - x^3$, $x^3 = (x^3+1) - 1 = (x+1)(x^2-x+1) - 1$. Then you "slowly" split the fraction to two fractions that can be simplified.


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