How do you integrate int (x^5 + 1)/(x^6 - x^4) using partial fractions?

1 Answer
Oct 12, 2016

int ((x^5 + 1)/(x^6 - x^4))dx=c_1logabsx+c_2/x-c_3/(2x^2)-c_4/(3x^3)+c_4log abs(x+1)+c_5 log abs(x-1)+c_0

Explanation:

(x^5 + 1)/(x^6 - x^4)=(x^5+1)/(x^4(x^2-1))=
=c_1/x+c_2/x^2+c_3/x^3+c_4/x^4+c_5/(x+1)+c_6/(x-1)

Then

int ((x^5 + 1)/(x^6 - x^4))dx=int(c_1/x+c_2/x^2+c_3/x^3+c_4/x^4+c_5/(x+1)+c_6/(x-1))dx=

=c_1logabsx+c_2/x-c_3/(2x^2)-c_4/(3x^3)+c_4log abs(x+1)+c_5 log abs(x-1)+C

The c_k determination follows.

Equating

(x^5 + 1)/(x^6 - x^4)=c_1/x+c_2/x^2+c_3/x^3+c_4/x^4+c_5/(x+1)+c_6/(x-1)

The condition to mantain the equality forall x in RR is the observation of

{(1 + c_4=0), (c_3=0), (c_2 - c_4=0), (c_1 - c_3=0), (c_2 - c_5 + c_6=0), (c_1 + c_5 + c_6-1=0):}

Solving we obtain

(c_1 = 0, c_2= -1, c_3= 0, c_4 = -1, c_5 = 0, c_6= 1)