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

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Answer 1 · 2017-12-20T06:33:27

#int(x^2-2x-1)/((x-1)^2(x^2+1))dx=ln|x-1|+1/(x-1)-1/2ln(x^2+1)+tan^(-1)x+c#

Let us first convert #(x^2-2x-1)/((x-1)^2(x^2+1))# to partial fractions. For this let

#(x^2-2x-1)/((x-1)^2(x^2+1))hArrA/(x-1)+B/(x-1)^2+(Cx+D)/(x^2+1)#

or #(x^2-2x-1)/((x-1)^2(x^2+1))=(A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2)/((x-1)^2(x^2+1))#

or #(x^2-2x-1)=A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2#

Putting #x=1# in this we get #2B=-2# or #B=-1#

Comparing coefficient of highest degree i.e. #x^3#, we get #A+C=0#,

comparing constant term we get #-A+B+D=-1# or #-A+D=0# i.e. #A=D#

and comparing coefficient of #x#, #-2=A-2D+C# i.e. #C-A=-2#

as #A+C=0#, we get #C=-1# and #A=1# and hence

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

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

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

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

= #ln|x-1|+1/(x-1)-1/2ln(x^2+1)+tan^(-1)x+c#