Solve $ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $ $$ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $$
I don't know about decomposition of fractions a lot,but I know that it is method which I have to use in this example. Please, help me. 

$$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2} $$
$$ \frac{x-1}{ x^2(x+1)^2} = A + \frac{Bx+C}{x^2}(x-2) + \frac{D(x+1)+E}{(x+1)^2}(x-2). $$
$x\to 2: A = \frac1{36}.$
$$ \frac{x-1}{(x-2)(x+1)^2} = \frac A{x-2}x^2+Bx+C + \frac{D(x+1)+E}{(x+1)^2}x^2 $$
$x\to 0: C = \frac12.$
$$ \frac{x-1}{(x-2)x^2} = \frac A{x-2}(x+1)^2+\frac{Bx+C}{x^2}(x+1)^2 + D(x+1)+E$$
$x\to -1: E = \frac23.$
$$ \frac{x-1}{(x-2)x(x+1)^2} = \frac A{x-2}x+\frac{Bx+C}{x} + \frac{D(x+1)+E}{(x+1)^2}x $$
$x\to\infty: A + B + D = 0.\quad (1)$
$$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2}$$
$x\to 1: -A+B+C+\frac{2D+E}4 = 0. \quad (2)$
$$(1)-(2): 2A-C+ \frac D2-\frac E4 = 0,$$
$$D=-4A+2C+\dfrac E2 = -\frac19+1+\frac13 = \frac{11}9$$
$B=-(D+A) = -(\frac{11}9+\frac1{36}) = -\frac 54$ 
