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#