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$$\frac{1-7x}{(1-x)^2(1+x)} = \frac{1-7x}{(1-x^2)(1-x)} = \frac{A}{1-x}+\frac{Bx+C}{1-x^2}$$ implies $1-7x = A-Ax^2+Bx-Bx^2+C-Cx$

Equating $x^2,x$ terms and constant on both sides, we get $$A+B = 0,\quad B-C=-7,\quad A+C=1$$ Therefore $$A+C=7,\quad A+C=1.$$

I am wondering how to resolve it into partial fractions to get the integration. I am not sure if I missed something, though it looks silly.

Thanks for any help.

Robert Z
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Santosh
  • 27

2 Answers2

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Another hint (a bit ad hoc): Solving $D$ and $E$ from $$D(x-1)+E(x+1)=7x-1$$ gives $D=4,E=3$. This allows you to rewrite $$ \frac{1-7x}{(1-x)^2(1+x)}=\frac{4(1-x)-3(x+1)}{(1-x)^2(1+x)}=\frac4{(1-x)(1+x)}-\frac3{(1-x)^2}. $$ Would this be easier?

Caveat: Only available because the numerator happened to be linear.

Jyrki Lahtonen
  • 133,153
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Your decomposition is not correct and it gives you an unsolvable system of constraints for the coefficients. Note that the multiplicity of the factor $(1-x)$ at the denominator is TWO. Hence the partial fraction decomposition should be $$\frac{1-7x}{(1-x)^2(1+x)} = \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+x} = \frac{(C-A)x^2 + (B-2C) x +(A+B+C)}{(1-x)^2(1+x)} .$$ Can you take it from here?

Robert Z
  • 145,942