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intercepts of 2log_{2}(2+x)-1

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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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laplace transformation (x+1)cos(x)

Solution

−−s²+1(s²+1)² +ss²+1

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L{cos(x)(x+1)}

Schreibe (x+1)cos(x) um: xcos(x)+cos(x)

=L{xcos(x)+cos(x)}

Verwende die lineare Eigenschaft der Laplace-Transformation:
Für Funktionen f(t), g(t) und Konstanten a, b: L{a·f(t)+b·g(t)}=a·L{f(t)}+b·L{g(t)}

=L{xcos(x)}+L{cos(x)}

L{xcos(x)}: −−s²+1(s²+1)²

L{cos(x)}: ss²+1

=−−s²+1(s²+1)² +ss²+1

Solve instead

inverse laplace transformation (x+1)cos(x)

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

intercepts of 2log_{2}(2+x)-1

Solution

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