How do you integrate (x^2+2x-1) / (x^3 - x)x2+2x1x3x?

1 Answer
Jun 20, 2016

int(x^2+2x-1)/(x^3-x)dx=lnx-ln(x+1)+ln(x-1)+cx2+2x1x3xdx=lnxln(x+1)+ln(x1)+c

Explanation:

As (x^2+2x-1)/(x^3-x)=(x^2+2x-1)/(x(x+1)(x-1))x2+2x1x3x=x2+2x1x(x+1)(x1), let us convert them into partial fractions

(x^2+2x-1)/(x^3-x)hArrA/x+B/(x+1)+C/(x-1)x2+2x1x3xAx+Bx+1+Cx1 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))x2+2x1x3x=A(x21)+B(x2x)+C(x2+x)x(x+1)(x1) or

(x^2+2x-1)/(x^3-x)=((A+B+C)x^2+(-B+C)x-A)/(x(x+1)(x-1))x2+2x1x3x=(A+B+C)x2+(B+C)xAx(x+1)(x1) i.e.

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

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

(x^2+2x-1)/(x^3-x)=1/x-1/(x+1)+1/(x-1)x2+2x1x3x=1x1x+1+1x1 and

int(x^2+2x-1)/(x^3-x)dx=int[1/x-1/(x+1)+1/(x-1)]dxx2+2x1x3xdx=[1x1x+1+1x1]dx

= lnx-ln(x+1)+ln(x-1)+clnxln(x+1)+ln(x1)+c