intervalo x^3+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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substituição ∫ 2x²x²−x dx

Solution

2(x−1+ln|2x−2|)+C

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∫ 2x²x²−x dx

Remover a constante: ∫a·f(x)dx=a·∫f(x)dx

=2· ∫ x²x²−x dx

Completar o cuadrado x²−x: (x−12 )²−14

=2· ∫ x²(x−12 )²−14 dx

Aplicar integração por substituição: ∫ 2u+12u−1 du

=2· ∫ 2u+12u−1 du

Expandir 2u+12u−1 : 2u2u−1 +12u−1

Aplicar a regra da soma: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=2(∫ 2u2u−1 du+∫ 12u−1 du)

∫ 2u2u−1 du=u−12 +12 ln|2u−1|

∫ 12u−1 du=12 ln|2u−1|

=2(u−12 +12 ln|2u−1|+12 ln|2u−1|)

Substituir na equação u=x−12

=2(x−12 −12 +12 ln|2(x−12 )−1|+12 ln|2(x−12 )−1|)

Simplificar 2(x−12 −12 +12 ln|2(x−12 )−1|+12 ln|2(x−12 )−1|): 2(x−1+ln|2x−2|)

=2(x−1+ln|2x−2|)

Adicionar uma constante à solução

=2(x−1+ln|2x−2|)+C

Verify Result

=2(x−1+ln|2x−2|)+C

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

intervalo x^3+1

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intervalo x^3+1

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