How do you integrate #(x^2+x-1)/(x(x^2-1))# using partial fractions?

2 Answers
Narad T.
Dec 5, 2016

The answer is #=ln(∣x∣)-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C#

Explanation:

We use

#a^2-b^2=(a+b)(a-b)#

So, the denominator is

#x(x^2-1)=x(x+1)(x-1)#

The decomposition in partial fractions is

#(x^2+x-1)/(x(x+1)(x-1))=A/x+B/(x+1)+C/(x-1)#

#(A(x+1)(x-1)+B(x)(x-1)+C(x)(x+1))/(x(x+1)(x-1))#

So,

#x^2+x-1=A(x+1)(x-1)+B(x)(x-1)+C(x)(x+1)#

Let, #x=0#,#=>#, #-1=-A#, #=>#, #A=1#

Let #x=1#, #=>#, #1=2C#, #=>#, #C=1/2#

Let #x=-1#,#=>#, #-1=2B#, #=>#, #B=-1/2#

So,

#(x^2+x-1)/(x(x+1)(x-1))=1/x+(-1/2)/(x+1)+(1/2)/(x-1)#

So,

#int((x^2+x-1)dx)/(x(x+1)(x-1))=intdx/x+int(-1/2dx)/(x+1)+int(1/2dx)/(x-1)#

#=ln(∣x∣)-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C#

Roy E.
Dec 5, 2016

Narad is correct. Alternatively use the cover-up rule to get #A#, #B# and #C# directly.

Explanation:

To get #A# "cover up" the #x# in the denominator and replace all other #x#'s with value of #x# which makes the covered-up term zero, which in this case #x=0#, so #A=(0^2+0-1)/((0+1)(0-1))=1#
and similarly #B# and #C# by putting #x=1# and #x=-1#.

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