Cauchy's integral formula - Self teaching: $\int_C\frac{5z^4-3z^3+2i}{(z-1)^3}dz $

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Hi I'm trying to teach myself Complex analysis and can't find many good examples online. When looking at a past exam paper I stumbled across this question:

Evaluate $$\ \int_C \frac {5z^4-3z^3+2i} {(z-1)^3} dz $$

Where C is any simple closed curve in a counterclockwise direction with z=1 inside C

Any help you can offer would really help out! Thanks

EDIT: $$ f^{(n)}(a) = \frac{2\pi i}{(n-1)!} \oint_C \frac{f(z)}{(z-a)^{n}}\, dz $$ with $$ f(z)=5z^4-3z^3+2i, \quad a=1, \quad n=3. $$ I get the answer to be: $$ \frac{2 \pi i}{2!}f^{(2)}(1)=\pi i (60-18)=42 \pi i $$

Think i may have done this wrong as it seems like an unusually large number?

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Hint. One may use Cauchy's integral formula $$ f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}}\, dz $$ with $$ f(z)=5z^4-3z^3+2i, \quad a=1, \quad n=2. $$