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The Laplace Transfor of a function $x(t)$ is given by 

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\\[X(s)=
\\frac{\\simplify{{A} + {B}}s^2 + \\simplify{{A}*{b1}*2 +{B}*{a1} + {C}}s + \\simplify{{A}*{c1}+{C}*{a1}}}{s^3 +\\simplify{{a1}+{{b1}*2}}s^2 +\\simplify{{c1}+{{a1}*{b1}*2}}s + \\simplify{{a1}*{c1}}}.\\]

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a) Observe that 

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\\[\\frac{\\simplify{{A}+{B}}s^2+\\simplify{{A}*{b1}*2 + {B}*{a1} + {C}}s + \\simplify{{A}*{c1}  + {C}*{a1}}}{s^3 +\\simplify{{a1}+{{b1}*2}}s^2 +\\simplify{{c1}+{{a1}*{b1}*2}}s + \\simplify{{a1}*{c1}}} =\\frac{\\var{A}s^2+\\simplify{{A}*{b1}*2}s + \\simplify{{A}*{c1}} +\\var{B}s^2 + \\simplify{{B}*{a1}}s +\\var{C}s + \\simplify{{C}*{a1}}}{s^3 +\\simplify{{a1}+{{b1}*2}}s^2 +\\simplify{{c1}+{{a1}*{b1}*2}}s + \\simplify{{a1}*{c1}}}= \\\\[6mm]
=\\frac{\\var{A}(s^2+\\simplify{{b1}*2}s+\\var{c1}) +(s+\\var{a1})(\\var{B}s+\\var{C})}{(s+\\var{a1})(s^2+\\simplify{{b1}*2}s+\\var{c1})} =\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\]



b)
We established 
\\[X(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\]

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in the previous part.

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By Completing the square we get:

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\\[X(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}.\\]

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Making it further resemble a transform in the Laplace Transfrom table we need to further manipulate the numerator of the second fraction, and then decompose it fruther:

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\\[X(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}(s+\\var{b1})-\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} = \\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}(s+\\var{b1})}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} -\\frac{\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}\\cdot\\frac{\\sqrt{\\simplify{{c1}-{b1}^2}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}.\\]

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Now, reading from the Fourier Transform table:

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\\[x(t)=\\var{A}e^{\\var{a1}t}+\\var{B}e^{-\\var{b1}t}\\cos\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)+\\frac{-\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}e^{-\\var{b1}t}\\sin\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)\\]

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Find the partial fraction decomposition of $X(s)$.

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$X(s) = $ 

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[[0]]

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Write down the inverse Laplace transform

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\\(x(t)=\\) [[0]]

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