end point (-9,4), (-13/2 ,6)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

simplify solve for inverse tangent line

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Steps Examples

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laplace transform (2x−5)e^x

Solution

2(s−1)² −5s−1

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Solution steps

One step at a time

L{e^x(2x−5)}

Apply transform rule: if L{f(t)}=F(s) then L{e^atf(t)}=F(s−a)

For (2x−5)e^x: f(x)=(2x−5), a=1

=L{(2x−5)}(s−1)

L{(2x−5)}: 2s² −5s

=2(s−1)² −5(s−1)

=2(s−1)² −5s−1

Solve instead

inverse laplace (2x−5)e^x

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Notebook

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

end point (-9,4), (-13/2 ,6)

Solution

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