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 $int(x^2-2x-1)/((x-1)^2(x^2+1))dx$

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

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

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

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