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integral of sin^5(x)cos^2(x)

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

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

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paridad (z−8)²z²−64

Solution

No es par ni impar

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∫ x⁴−x²+2x²(x−1) dx

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

=∫ x+1−2x −2x² +2x−1 dx

Aplicar la regla de la suma: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ xdx+∫ 1dx−∫ 2x dx−∫ 2x² dx+∫ 2x−1 dx

∫ xdx=x²2

∫ 1dx=x

∫ 2x dx=2ln|x|

∫ 2x² dx=−2x

∫ 2x−1 dx=2ln|x−1|

=x²2 +x−2ln|x|−(−2x )+2ln|x−1|

Simplificar

=x²2 +x−2ln|x|+2x +2ln|x−1|

Agregar una constante a la solución

=x²2 +x−2ln|x|+2x +2ln|x−1|+C

Solve instead

domínio (z−8)²z²−64

rango (z−8)²z²−64

inversa (z−8)²z²−64

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

integral of sin^5(x)cos^2(x)

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