Question 357825
Part a is correct but why not {{{-3}}} instead of {{{(1.5/(-0.5))}}}??
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Part b denominator can be factored to {{{(x+1)(x^2-1)=(x+1)(x-1)(x+1)=(x-1)(x+1)^2}}}
So look for a partial fraction expansion of,
{{{ 1/((x+1)(x^2-1)) =A/(x-1)+B/(x+1)+C/(x+1)^2}}}