<s>[INST] Find inverse $z$-transform of $\frac{5}{z^{2}-z-6}$ How can I find inverse z transform of $$X(z)=\frac{5}{z^{2}-z-6}$$ 
What I did:
first i factored denominator and i got (z+2)(z-3), now we get A(-2^{n}) + b(3^{n}). To get A and B i used Partial Fraction Decomposition and got A=-1 and B=1.
But Wolframalpha gave me another answer! $$=\frac{1}{6}(3(-2^{n}+2*3^{n}))$$  what i did wrong?
 [/INST] Doing partial fractions as per usual,
$$\frac{5}{(z+2)(z-3)}= \frac{A}{z+2} + \frac{B}{z-3} \implies 5 = A(z-3)+ B(z+2).$$
Equating coefficients, we see that
$$A+B=0 \mbox{ and } 2B-3A=5.$$
This linear system has the unique solution $A=-1$, $B=1$. Therefore, we see that
$$\frac{5}{(z+2)(z-3)}=\frac{1}{z-3} - \frac{1}{z+2}.$$
So the poles of $X(z)$ are encircled, choose $C$ to be $\{z\in\mathbb{C}:|z|=4\}$ with standard orientation. 
Therefore, by definition, the inverse $Z$-transform of $X(z)$ is
\begin{align*}
\frac{1}{2\pi j} \int_C X(z)z^{n-1} dz &= \frac{1}{2\pi j} \int_C \left(\frac{1}{z-3} - \frac{1}{z+2}\right)z^{n-1} dz \\
&= \frac{1}{2\pi j} \left(\int_C \frac{z^{n-1}}{z-3}dz - \int_C\frac{z^{n-1}}{z+2} dz\right). \\
\end{align*}
We then apply Cauchy's Integral Formula to get
$$\frac{1}{2\pi j} \left(\int_C \frac{z^{n-1}}{z-3}dz - \int_C\frac{z^{n-1}}{z+2} dz\right)=3^{n-1} - (-2)^{n-1}.$$
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