How do you use partial fraction decomposition to decompose the fraction to integrate $(x^5 + 1)/(x^6 - x^4)$ ?

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Answer 1 · 2015-08-01T17:49:08

$(x^5 + 1)/(x^6 - x^4) = -1/x^2-1/x^4+1/(x-1)$

Factor the denominator:

$x^6-x^4 - x^4(x^2-1) = x^4(x+1)(x-1)$

Perhaps we know that $x^5+1$ can be factored, or perhaps we notice that $-1$ is a zero of the numerator, so $x-1$ is a factor.

In any case, we can write:

$(x^5 + 1)/(x^6 - x^4) = ((x+1)(x^4-x^3+x^2-x+1))/(x^4(x+1)(x-1))$

So

$(x^5 + 1)/(x^6 - x^4) = (x^4-x^3+x^2-x+1)/(x^4(x-1))$

$A/x+B/x^2+C/x^3+D/x^4+E/(x-1) = (x^4-x^3+x^2-x+1)/(x^4(x-1))$

Leads to:

$(Ax^4-Ax^3+Bx^3-Bx^2+Cx^2-Cx+Dx-D+Ex^4)/(x^4(x-1))$

$= (x^4-x^3+x^2-x+1)/(x^4(x-1))$

So
$A+E = 1$
$-A+B=-1$
$-B+C = 1$
$-C+D=-1$
$-D = 1$

So $D=-1$ and,working back through the list:

$C =0" "$ , $B = -1" "$ , $A = 0" "$ , and $" "E = 1$ .

But what if? What if we didn't notice that the fraction can be reduced?

It's OK, just a bit longer.

We would get:

$(x^5 + 1)/(x^6 - x^4) = (x^5+1)/(x^4(x+1)(x-1))$

$A/x+B/x^2+C/x^3+D/x^4+E/(x-1) +F/(x+1) = (x^5+1)/(x^4(x+1)(x-1))$

In this case we get one more variable (and one more equation).

When we solve, we'll get the same $A, B, C, D, "and " E$ and we will get $F=0$

How do you use partial fraction decomposition to decompose the fraction to integrate (x^5 + 1)/(x^6 - x^4)(x^5 + 1)/(x^6 - x^4)?