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codominio sin^2(x)

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∫ x+x³+x⁵1−x²+x⁴ dx

Solution

12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))+C

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Solution steps

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∫ x+x³+x⁵1−x²+x⁴ dx

Applicare la sostituzione u: ∫ 1+u+u²2(u²+1−u) du

=∫ 1+u+u²2(u²+1−u) du

Porta fuori la costante: ∫a·f(x)dx=a·∫f(x)dx

=12 · ∫ 1+u+u²u²+1−u du

Espandi 1+u+u²u²+1−u : 1u²+1−u +uu²+1−u +u²u²+1−u

Applica la Regola della Somma: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=12 (∫ 1u²+1−u du+∫ uu²+1−u du+∫ u²u²+1−u du)

∫ 1u²+1−u du=2√3 arctan(2u−1√3 )

∫ uu²+1−u du=12 ln|4u²−4u+4|+1√3 arctan(2u−1√3 )

∫ u²u²+1−u du=12 ln|4u²−4u+4|−1√3 arctan(2u−1√3 )+u

=12 (2√3 arctan(2u−1√3 )+12 ln|4u²−4u+4|+1√3 arctan(2u−1√3 )+12 ln|4u²−4u+4|−1√3 arctan(2u−1√3 )+u)

Sostituire indietro u=x²

=12 (2√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|+1√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|−1√3 arctan(2x²−1√3 )+x²)

Semplificare 12 (2√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|+1√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|−1√3 arctan(2x²−1√3 )+x²): 12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))

=12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))

Aggiungi una costante alla soluzione

=12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))+C

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

codominio sin^2(x)

Solution

Solution steps

Number Line

Graph

1	−1	0
1	0	1
0	−1	1

1	1	1	2
1	0	1	−1
0	−1	−2	1
0	2	0	2