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inverse of 1/(x^2)

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

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∫ 2x+1(x+5)³ dx

Solution

−2x+5 +92(x+5)² +C

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∫ 2x+1(x+5)³ dx

Appliquer l'intégration par subsitution: ∫ 2u−9u³ du

=∫ 2u−9u³ du

Développer 2u−9u³ : 2u² −9u³

=∫ 2u² −9u³ du

Appliquer la règle de la somme: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ 2u² du−∫ 9u³ du

∫ 2u² du=−2u

∫ 9u³ du=−92u²

=−2u −(−92u² )

Remplacer u=x+5

=−2x+5 −(−92(x+5)² )

Simplifier

=−2x+5 +92(x+5)²

Ajouter une constante à la solution

=−2x+5 +92(x+5)² +C

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

inverse of 1/(x^2)

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