Question

$(b - 1) {}^{2} ( \frac{1}{b {}^{2} - 2b + 1} + \frac{1}{b {}^{2} - 1 } ) + \frac{2}{b + 1} =$
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Answer 1

$\tt (b-1)^2\bigg(\dfrac{1}{b^2-2b+1}+\dfrac{1}{b^2-1}\bigg)+\dfrac{2}{b+1}=(b-1)^2\bigg(\dfrac{1}{(b-1)^2}+\\ \\ \\ +\dfrac{1}{(b-1)(b+1)}\bigg)+\dfrac{2}{b+1}=1+\dfrac{b-1}{b+1}+\dfrac{2}{b+1}=1+\dfrac{b-1+2}{b+1}=\\ \\ \\ =1+\dfrac{b+1}{b+1}=1+1=2$