Evaluate $\int \frac{dx}{x(x-1)^3(x-2)^2}$ Evaluate $$I=\int \frac{dx}{x(x-1)^3(x-2)^2}$$ without using tedious partial fractions.
My Try: 
we have $$1=\left((x-1)-(x-2)\right)^2$$ So
 $$=\int \frac{\left((x-1)-(x-2)\right)^2dx}{x(x-1)^3(x-2)^2}$$ So
$$I=\int \frac{dx}{x(x-1)(x-2)^2}+\int \frac{dx}{x(x-1)^3}-2\int \frac{dx}{x(x-1)^2(x-2)} $$
any clue here?

You can reduce the labour a little by writing
$\dfrac{1}{x(x-1)^3(x-2)^2} = \dfrac{(1-x+x^2)+x(x-1)}{x(x-1)^3(x-2)^2}$
$=\dfrac{x(x-2)^2-(x-1)^3}{x(x-1)^3(x-2)^2} + \dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3} -  \dfrac{1}{x(x-2)^2}+\dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3}- \dfrac{1}{x(x-2)^2}+ \dfrac{1}{(x-1)^2}+\dfrac{1}{(x-2)^2}+2 \left[\dfrac{1}{x-1} - \dfrac{1}{x-2} \right]$
If you wish you can further decompose $\dfrac{1}{x(x-2)^2}= \dfrac{1}{(x-2)^2}+\dfrac{1}{2} \left[\dfrac{1}{x} - \dfrac{1}{x-2} \right]$
This expression can be readily integrated.
