<s>[INST] 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
 [/INST] Hint
Write the integrand as: $$\frac{3x-2}{x^2(x-1)}=\frac{Ax+B}{x^2}+\frac{C}{x-1}$$ in which $A,B$ and $C$ is unknown constants and so find the proper values for them. I think via this way you can find them easier than the way you noted. Indeed, $$\frac{Ax+B}{x^2}+\frac{C}{x-1}=\frac{(A+C)x^2+x(B-A)-B}{x^2(x-1)}$$ and therefore you get $$A+C=0, ~~B-A=3,~~-B=-2$$ and so...
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