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absolute extreme points of y=(x^2+x+1)/x

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

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∫ √1+4x²dx

Solution

14 (2x√4x²+1+ln|2x+√1+4x²|)+C

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Solution steps

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∫ √1+4x²dx

Appliquer la subsitution trigonométrique: ∫ 12 sec³(u)du

=∫ 12 sec³(u)du

Extraire la constante: ∫a·f(x)dx=a·∫f(x)dx

=12 · ∫ sec³(u)du

Appliquer une réduction intégrale

=12 (sec²(u)sin(u)2 +12 · ∫ sec(u)du)

∫ sec(u)du=ln|tan(u)+sec(u)|

=12 (sec²(u)sin(u)2 +12 ln|tan(u)+sec(u)|)

Remplacer u=arctan(2x)

=12 (sec²(arctan(2x))sin(arctan(2x))2 +12 ln|tan(arctan(2x))+sec(arctan(2x))|)

Simplifier 12 (sec²(arctan(2x))sin(arctan(2x))2 +12 ln|tan(arctan(2x))+sec(arctan(2x))|): 14 (2x√4x²+1+ln|2x+√1+4x²|)

=14 (2x√4x²+1+ln|2x+√1+4x²|)

Ajouter une constante à la solution

=14 (2x√4x²+1+ln|2x+√1+4x²|)+C

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

absolute extreme points of y=(x^2+x+1)/x

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