Find partial fractions of, $$\frac{3x^2-2x+1}{x^2(1-x^2)}$$

Hence deduce partial fractions of $$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$$

My Try

I was able to do the first part,$$\frac{3x^2-2x+1}{x^2(1-x^2)}=\frac1{x^2}-\frac2x+\frac3{x+1}-\frac1{x-1}$$

But I find it difficult to do the second part. Should I use a substitute or any other method? Please give me a hint to work this out. Thank you very much!


Let $X=x-1$ then $$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}=-\frac{3(X+1)^2-8(X+1)+6}{(X+1)X^2(X-1)}= -\frac{3X^2-2X+1}{X^2(X^2-1)}$$ Now you can use the first part.


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