Using partial fractions for "F(s) = \\dfrac{8s^{2}-12s+144}{s(s^{2}+36)}" ,
"\\begin{aligned}\n\\dfrac{8s^{2}-12s+144}{s(s^{2}+36)} &= \\dfrac{A}{s}+\\dfrac{Bs+C}{s^{2}+36}\\\\\n8s^{2}-12s+144 & = A(s^{2}+36) + (Bs+C)s\\\\\n8s^{2}-12s+144 & =(A+B)s^{2}+Cs+36A\n\\end{aligned}"
Comparing the coefficients of "s^{2}, s" and constant terms, we get
"\\begin{aligned}\nA+B &= 8\\\\\nC &= -12\\\\\nA = 4 &\\Rightarrow B = 4\n\\end{aligned}"
Therefore,
"\\begin{aligned}\nL^{-1}\\{F(s)\\} &= L^{-1}\\left\\{\\dfrac{4}{s} + \\dfrac{4s-12}{s^{2}+36} \\right\\}\\\\\n&=L^{-1}\\left\\{\\dfrac{4}{s}\\right\\}+L^{-1}\\left\\{\\dfrac{4s}{s^{2}+36}\\right\\}-L^{-1}\\left\\{\\dfrac{12}{s^{2}+36}\\right\\}\\\\\n&=4L^{-1}\\left\\{\\dfrac{1}{s}\\right\\}+4L^{-1}\\left\\{\\dfrac{s}{s^{2}+36}\\right\\}-2L^{-1}\\left\\{\\dfrac{6}{s^{2}+36}\\right\\}\\\\\nL^{-1}\\{F(s)\\}&=4+4\\cos(6t)-2\\sin(6t) \n\\end{aligned}"
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