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)#