Q.Resolve $\dfrac{x^2 + 5x + 7}{(x - 3)^3}$ into partial fractions.

Putting $x=t+3$ gives numerator $t^2+11t+31$, so the fractions are $\dfrac{1}{x-3}+\dfrac{11}{(x-3)^2}+\dfrac{31}{(x-3)^3}$.

Assume $\dfrac{x^2+5x+7}{(x-3)^3} = \dfrac{A}{x-3} + \dfrac{B}{(x-3)^2} + \dfrac{C}{(x-3)^3}$.

A quick way: substitute $t = x - 3$, so $x = t + 3$.

Numerator $= (t+3)^2 + 5(t+3) + 7 = t^2 + 6t + 9 + 5t + 15 + 7 = t^2 + 11t + 31$.

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