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

Question

1207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-25T14:29:46

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

Reminder

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

Let's perform the decomposition into partial fractions

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

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

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

The denominators are the same, compare the numerators

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

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

Let #x=1#, #=>#, #1=4D#, #=>#, #D=1/4#

Let #x=-1#, #=>#, #1=-4B#, #=>#, #B=-1/4#

Coefficients of #x^4#

#0=A+C+E#, #=>#, #C+E=-A=-1#

Coefficients of #x^3#

#0=B+D-C+E#, #=>#, #B+D=C-E=0#

#C=E=-1/2#

Therefore,

#1/(x(x^2-1)^2)=1/x+(-1/4)/(x+1)^2+(-1/2)/(x+1)+(1/4)/(x-1)^2+(-1/2)/(x-1)#

So,

#int(dx)/(x(x^2-1)^2)=int(dx)/x-int(1/4dx)/(x+1)^2-int(1/2dx)/(x+1)+int(1/4dx)/(x-1)^2-int(1/2dx)/(x-1)#

#=ln(|x|)+1/(4(x+1))-1/2ln(|x+1|)-1/(4(x-1))-1/2ln(|x-1|)+C#