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improper fraction 23/7+85/7

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

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∫ √x²+2dx

Solution

12 x√x²+2+ln|1√2 x+√12 (2+x²)|+C

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∫ √x²+2dx

Appliquer la subsitution trigonométrique: ∫ 2sec³(u)du

=∫ 2sec³(u)du

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

=2· ∫ sec³(u)du

Appliquer une réduction intégrale

=2(sec²(u)sin(u)2 +12 · ∫ sec(u)du)

∫ sec(u)du=ln|tan(u)+sec(u)|

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

Remplacer u=arctan(1√2 x)

=2(sec²(arctan(1√2 x))sin(arctan(1√2 x))2 +12 ln|tan(arctan(1√2 x))+sec(arctan(1√2 x))|)

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

=12 x√x²+2+ln|1√2 x+√12 (2+x²)|

Ajouter une constante à la solution

=12 x√x²+2+ln|1√2 x+√12 (2+x²)|+C

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

improper fraction 23/7+85/7

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improper fraction 2*3/7+8*5/7

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improper fraction 23/7+85/7