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invers x/(x+1)

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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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trennbar dxdt =5x−3

Solution

x=e^5t+c₁5 +35

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L⁻¹{xx²−6x+8 }

Ermittle den Partialbruch von xx²−6x+8 : −1x−2 +2x−4

=L⁻¹{−1x−2 +2x−4 }

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:
Für Funktionen f(s), g(s) und Konstanten a, b: L⁻¹{a·f(s)+b·g(s)}=a·L⁻¹{f(s)}+b·L⁻¹{g(s)}

=−L⁻¹{1x−2 }+L⁻¹{2x−4 }

L⁻¹{1x−2 }: e^2t

L⁻¹{2x−4 }: 2e^4t

=−e^2t+2e^4t

Order of operations multi-step without parentheses Examples

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

invers x/(x+1)

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