integral of x/(x^2-25)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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inverse laplace (x−6)(5x²−6x+8)

Solution

15 e^3t5cos(√31t5 )−27e^3t5sin(√31t5 )5√31

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L⁻¹{(x−6)(5x²−6x+8) }

Expand x−65x²−6x+8 : 15 · x−35 (x−35 )²+3125 −2725 · 1(x−35 )²+3125

=L⁻¹{15 · x−35 (x−35 )²+3125 −2725 · 1(x−35 )²+3125 }

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)}

=15 L⁻¹{x−35 (x−35 )²+3125 }−2725 L⁻¹{1(x−35 )²+3125 }

L⁻¹{x−35 (x−35 )²+3125 }: e^3t5cos(√31t5 )

L⁻¹{1(x−35 )²+3125 }: e^3t55√31 sin(√31t5 )

=15 e^3t5cos(√31t5 )−2725 e^3t55√31 sin(√31t5 )

Refine 15 · e^3t5cos(√31t5 )−2725 · e^3t55√31 sin(√31t5 ): 15 e^3t5cos(√31t5 )−27e^3t5sin(√31t5 )5√31

=15 e^3t5cos(√31t5 )−27e^3t5sin(√31t5 )5√31

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

integral of x/(x^2-25)

Solution

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