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integral of x^5ln(x)

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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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intersecciones (x²+2x−1)· (2x²−3x+6)

Solution

X intersecta: (−1+√2, 0), (−1−√2, 0), Y intersecta: (0, −6)

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Solution steps

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L⁻¹{x⁴−x²+2x²(x−1) }

Tomar la fracción parcial de x⁴−x²+2x²(x−1) : x+1−2x −2x² +2x−1

=L⁻¹{x+1−2x −2x² +2x−1 }

Usar la propiedad de linealidad de la transformada inversa de Laplace:
Para las funciones f(s), g(s) y 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{1}−L⁻¹{2x }−L⁻¹{2x² }+L⁻¹{2x−1 }

L⁻¹{x}: δ′ (t)

L⁻¹{1}: δ(t)

L⁻¹{2x }: 2H(t)

L⁻¹{2x² }: 2t

L⁻¹{2x−1 }: 2e^t

=δ′ (t)+δ(t)−2H(t)−2t+2e^t

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

integral of x^5ln(x)

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