I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got : $\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions: $ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
Well I got A=1/3 and B=-1/3. as possible values. but that was not correct. I guess I missed something.
On
Instead of $\frac{Ax}{(x-1)^2}$ you should have $$\frac A{(x-1)^2}+\frac C{x-1}$$ I get $A=\frac13,B=\frac19,C=-\frac19$.
On
When the denominator of a fraction has an irreducible factor $\bigl(p(x)\bigr)^m$ with order of multiplicity $m>1$, its contribution to the decomposition into partial fractions is not $$\frac{A(x)}{\bigl(p(x)\bigr)^m} \qquad (\deg A(x)<\deg p(x)),$$ but $$\frac{A_1(x)}{p(x)\vphantom{\Big)}}+\frac{A_2(x)}{\bigl(p(x)\bigr)^2}+\dots+\frac{A_m(x)}{\bigl(p(x)\bigr)^m}\\(\deg A_1(x),\deg A_2(x),\dots,\deg A_m(x)<\deg p(x)) $$ Therefore here, as the irreducible factors have degree $1$, the decomposition has the form $$\frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}.$$
Addendum:
Here is the short way to determine the coefficients: multiply both sides of the above equality by $(x-1)^2(x+2)$ to obtain the relation $$x=(x-1)(x+2)A+(x+2)B+(x-1)^2C.$$ Setting $x=1$, then $x=-2$, yields instantly $B=\frac 13$ and $C=-\frac29$. To obtain $C$, observe that the r.h.s. must have no quadratic term, so $A+C=0$ and finally $A=\frac29$.
On
$$ I=\int \frac{x}{x^3-3x+2}dx$$ $$I= \int \frac {xdx}{(x-1)^2(x+2)}$$ Since $\frac {-1}{(x-1)^2}$ is an obvious derivative substitute $$u=\dfrac {1}{x-1} \implies -du=\dfrac {dx}{(x-1)^2}$$ The integral becomes: $$I=-\int \dfrac {u+1}{3u+1}du$$ Which is easier to integrate. $$u+1=u+\frac 1 3 +\frac 2 3$$
You don't have all the components in the decomposition. It should be
$$\frac {x}{(x-1)^2(x+2)} = \frac {Ax+C}{(x-1)^2} + \frac {B}{x+2} =\frac {\frac29x+\frac19}{(x-1)^2} - \frac {\frac29}{x+2}$$