We have:
int dx/(x(x^2-1)^2)
Multiply and divide the integrand function by x:
int dx/(x(x^2-1)^2) = int (xdx)/(x^2(x^2-1)^2
Substitute now: x^2 = t; 2xdx = dt
int dx/(x(x^2-1)^2) = 1/2 int (dt)/(t(t-1)^2)
Add and subtract t to the numerator:
int dx/(x(x^2-1)^2) = 1/2 int (1-t+t)/(t(t-1)^2)dt
using linearity:
int dx/(x(x^2-1)^2) = 1/2 int t/(t(t-1)^2)dt-1/2 int (t-1)/(t(t-1)^2)dt
and simplifying:
int dx/(x(x^2-1)^2) = 1/2 int 1/((t-1)^2)dt-1/2 int 1/(t(t-1))dt
The first integral can be resolved directly:
(1) int dx/(x(x^2-1)^2) = -1/2 1/(t-1)-1/2 int 1/(t(t-1))dt
For the second, we can divide in partial fractions using the same method:
int 1/(t(t-1))dt = int (1-t+t)/(t(t-1))dt = int (1-t)/(t(t-1))dt + int (tdt)/(t(t-1))
int 1/(t(t-1))dt = - int(dt)/t + int (dt)/(t-1)
int 1/(t(t-1))dt = -ln abst +lnabs(t-1)+C
Substituting in the expression (1) above:
int dx/(x(x^2-1)^2) = -1/2 1/(t-1) +1/2lnabs t -1/2 lnabs(t-1)
and undoing the variable substitution:
int dx/(x(x^2-1)^2) = -1/2 1/(x^2-1) +1/2lnx^2 -1/2 ln abs(x^2-1)
and finally, using the properties of logarithms:
int dx/(x(x^2-1)^2) = -1/2 1/(x^2-1) +ln(absx/sqrt abs(x^2-1))
