How do you find the indefinite integral of #int (x(x-2))/(x-1)^3#?

1 Answer
Dec 19, 2016

Substitute #u=x-1# to get #ln|x-1|+1/(2(x-1)^2)+C#

Explanation:

#int (x(x-2))/(x-1)^3dx#
Substitute #u=x-1#, #dx=du#:
#=int((u+1)(u-1))/u^3du#
#=int u^-1-u^-3dx#
#=ln|u|+1/2u^-2+C#
#=ln|x-1|+1/(2(x-1)^2)+C#

Whenever the denominator of the rational function is just a (positive) power of a (linear?) expression this will work, because it "pushes all the messy stuff upstairs".