How do you integrate $(x^2+2x-1) / (x^3 - x)$ ?

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Answer 1 · 2016-06-20T09:01:37

$int(x^2+2x-1)/(x^3-x)dx=lnx-ln(x+1)+ln(x-1)+c$

As $(x^2+2x-1)/(x^3-x)=(x^2+2x-1)/(x(x+1)(x-1))$ , let us convert them into partial fractions

$(x^2+2x-1)/(x^3-x)hArrA/x+B/(x+1)+C/(x-1)$ or

$(x^2+2x-1)/(x^3-x)=(A(x^2-1)+B(x^2-x)+C(x^2+x))/(x(x+1)(x-1))$ or

$(x^2+2x-1)/(x^3-x)=((A+B+C)x^2+(-B+C)x-A)/(x(x+1)(x-1))$ i.e.

$A+B+C=1$ , $-B+C=2$ and $A=1$ ,

which gives $C=1$ and $B=-1$ and hence

$(x^2+2x-1)/(x^3-x)=1/x-1/(x+1)+1/(x-1)$ and

$int(x^2+2x-1)/(x^3-x)dx=int[1/x-1/(x+1)+1/(x-1)]dx$

= $lnx-ln(x+1)+ln(x-1)+c$

How do you integrate (x^2+2x-1) / (x^3 - x)(x^2+2x-1) / (x^3 - x)?