integral of tan^4(x)sec^6(x)

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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inverse laplace x³−4x−10x²−x−6

Solution

δ′(t)+δ(t)+2e^−2t+e^3t

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L⁻¹{x³−4x−10x²−x−6 }

Take the partial fraction of x³−4x−10x²−x−6 : x+1+2x+2 +1x−3

=L⁻¹{x+1+2x+2 +1x−3 }

Use the linearity property of Inverse Laplace Transform:
For functions f(s), g(s) and constants a, b: L⁻¹{a·f(s)+b·g(s)}=a·L⁻¹{f(s)}+b·L⁻¹{g(s)}

=L⁻¹{x}+L⁻¹{1}+L⁻¹{2x+2 }+L⁻¹{1x−3 }

L⁻¹{x}: δ′ (t)

L⁻¹{1}: δ(t)

L⁻¹{2x+2 }: 2e^−2t

L⁻¹{1x−3 }: e^3t

=δ′ (t)+δ(t)+2e^−2t+e^3t

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

integral of tan^4(x)sec^6(x)

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