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inverse of (x^2+1)(x^2-1)

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

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implicit derivative dydx , y²(y²−4)=x²(x²−5)

Solution

dydx =x(2x²−5)2y(y²−2)

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Solution steps

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∫ cos⁵(x)sin²(x)dx

Rewrite using trig identities

=∫ (1−sin²(x))²cos(x)sin²(x)dx

Apply u-substitution: ∫ u²(1−u²)²du

=∫ u²(1−u²)²du

Expand u²(1−u²)²: u²−2u⁴+u⁶

=∫ u²−2u⁴+u⁶du

Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ u²du−∫ 2u⁴du+∫ u⁶du

∫ u²du=u³3

∫ 2u⁴du=2u⁵5

∫ u⁶du=u⁷7

=u³3 −2u⁵5 +u⁷7

Substitute back u=sin(x)

=sin³(x)3 −2sin⁵(x)5 +sin⁷(x)7

Add a constant to the solution

=sin³(x)3 −2sin⁵(x)5 +sin⁷(x)7 +C

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

inverse of (x^2+1)(x^2-1)

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