inverse of arccos(x)

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

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ddx

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

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inverse laplace 4x−7x²+5x

Solution

−75 H(t)+275 e^−5t

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Solution steps

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L⁻¹{4x−7x²+5x }

Take the partial fraction of 4x−7x²+5x : −75x +275(x+5)

=L⁻¹{−75x +275(x+5) }

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{75x }+L⁻¹{275(x+5) }

L⁻¹{75x }: 75 H(t)

L⁻¹{275(x+5) }: 275 e^−5t

=−75 H(t)+275 e^−5t

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

inverse of arccos(x)

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