Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$ To be shown that $\int_2^\infty{\dfrac{3x-2}{x^2(x-1)}}=1-\ln2$
My thought: $\dfrac{3x-2}{x^2(x-1)}=\dfrac{3x}{x^2(x-1)}-\dfrac{2}{x^2(x-1)}$
• $\dfrac{3x}{x^2(x-1)}=\dfrac{3}{x(x-1)}=\ldots=-\dfrac{3}{x}+\dfrac{3}{x-1}$
• $\dfrac{2}{x^2(x-1)}=\ldots=-\dfrac{2}{x^2}+\dfrac{2}{x-1}$
But developing the sum of the integrals of the above two gives ln of infinite in my results. I don't know what I am doing wrong. Can you please confirm the dots above? Although I have several times. Thanks a lot

HINT:
Using Partial Fraction Decomposition,
$$\dfrac{3x-2}{x^2(x-1)}=\frac A{x^2}+\frac Bx+\frac C{x-1}$$
$$\implies 3x-2=A(x-1)+Bx(x-1)+Cx^2=(B+C)x^2+x(A-B)-A$$
Comparing the constants in the above identity $A=2$
Comparing the  coefficients of $x,A-B=3\implies B=A-3=-1$
Comparing the  coefficients of $x^2,B+C=0\implies C=-B=1$
Alternatively, $$  3x-2=A(x-1)+Bx(x-1)+Cx^2$$
Putting $x=1$ in the above identity, $C=3\cdot 1-2=1$ 
Putting $x=0,-2=A(0-1)\implies A=2$
Comparing the  coefficients of $x^2,B+C=0\implies B=-C=-1$
