S [(x(x-2)/(x-1)^3]
I can't figure out what u should be.
first we need to do some manipulation (if we don't do it now, we'd end up doing it later, so let's get it over with).
Note that $\displaystyle \frac {x(x - 2)}{(x - 1)^3} = \frac {x^2 - 2x}{(x - 1)^3} = \frac {x^2 - 2x + 1 - 1}{(x - 1)^3} = \frac {x^2 - 2x + 1}{(x - 1)^3} - \frac 1{(x - 1)^3}$
thus, $\displaystyle \int \frac {x(x - 2)}{(x - 1)^3}~dx = \int \frac {x^2 - 2x + 1}{(x - 1)^3}~dx - \int \frac 1{(x - 1)^3}~dx$
now, for the first integral, use the substitution $\displaystyle u = (x - 1)^3$ for the second integral, use the substitution $\displaystyle t = x - 1$
Ah, i just saw it. an easier way is to let $\displaystyle u = x - 1$ from the beginning. that one substitution will finish the problem
you would end up with $\displaystyle \int \left( \frac 1u - \frac 1{u^3}\right)~du$, which should be easy for you
can you continue?