یکبار جواب 2a با سری و جواب b و c رو به صورت فقط ریاضی بدون توضیح برام پشت سر هم بنویس 

### جواب مسئله 2a با استفاده از سری تیلور:
\[
\cot^{-1}(s + 1) = \sum_{n=0}^{\infty} \frac{(-1)^n (s + 1)^{2n+1}}{2n+1}
\]
تبدیل لاپلاس معکوس:
\[
\mathcal{L}^{-1}\left\{\cot^{-1}(s + 1)\right\} = \mathcal{L}^{-1}\left\{\sum_{n=0}^{\infty} \frac{(-1)^n (s + 1)^{2n+1}}{2n+1}\right\} = \sum_{n=0}^{\infty} \frac{(-1)^n \mathcal{L}^{-1}\left\{(s + 1)^{2n+1}\right\}}{2n+1}
\]
\[
\mathcal{L}^{-1}\left\{(s + 1)^{2n+1}\right\} = t^{-2n-2} e^{-t}
\]
\[
f(t) = \sum_{n=0}^{\infty} \frac{(-1)^n t^{-2n-2} e^{-t}}{2n+1}
\]

### جواب مسئله 2b:
\[
\mathcal{L}^{-1}\left\{\frac{e^{-s}}{(s^2 + 4)^2}\right\} = u(t-1) \frac{(t-1) \sin(2(t-1))}{4}
\]

### جواب مسئله 2c:
\[
\frac{s - 6}{s^3 - 2s^2 + 6s} = \frac{-1}{s} + \frac{s + 1}{s^2 - 2s + 6}
\]
\[
\frac{s + 1}{s^2 - 2s + 6} = \frac{s - 1 + 2}{(s-1)^2 + 5} = \frac{s-1}{(s-1)^2 + 5} + \frac{2}{(s-1)^2 + 5}
\]
\[
\mathcal{L}^{-1}\left\{\frac{s-1}{(s-1)^2 + 5}\right\} = e^t \cos(\sqrt{5}t)
\]
\[
\mathcal{L}^{-1}\left\{\frac{2}{(s-1)^2 + 5}\right\} = \frac{2}{\sqrt{5}} e^t \sin(\sqrt{5}t)
\]
\[
\mathcal{L}^{-1}\left\{\frac{s - 6}{s^3 - 2s^2 + 6s}\right\} = -1 + e^t \cos(\sqrt{5}t) + \frac{2}{\sqrt{5}} e^t \sin(\sqrt{5}t)
\]

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