points of d'intersection (x^2-4)(5x+16)

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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 expand factor rationalize

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∫ sec⁵(x)tan⁵(x)dx

Solution

sec⁹(x)9 −2sec⁷(x)7 +sec⁵(x)5 +C

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Solution steps

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L⁻¹{x²−4x−2 }

Prendre la fraction partielle de x²−4x−2 : x+2

=L⁻¹{x+2}

Utiliser la propriété linéaire de la transformée inverse de Laplace :
Pour les fonctions f(s), g(s) et les constantes a, b: L⁻¹{a·f(s)+b·g(s)}=a·L⁻¹{f(s)}+b·L⁻¹{g(s)}

=L⁻¹{x}+L⁻¹{2}

L⁻¹{x}: δ′ (t)

L⁻¹{2}: 2δ(t)

=δ′ (t)+2δ(t)

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

points of d'intersection (x^2-4)(5x+16)

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