vertices (-3x^2-2x+6)/5

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

☐²

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ddx

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

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∫ √25+x²dx

Solution

25(150 x√25+x²+12 ln(|x+√25+x²|5 ))+C

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∫ √25+x²dx

Apply Trigonometric Substitution: ∫ 25sec³(u)du

=∫ 25sec³(u)du

∫a·f(x)dx=a·∫f(x)dx :قم بإخراج الثابت

=25· ∫ sec³(u)du

فعّل صيغة التكامل بالتخفيض

=25(sec²(u)sin(u)2 +12 · ∫ sec(u)du)

∫ sec(u)du=ln|tan(u)+sec(u)|

=25(sec²(u)sin(u)2 +12 ln|tan(u)+sec(u)|)

u=arctan(15 x) استبدل مجددًا

=25(sec²(arctan(15 x))sin(arctan(15 x))2 +12 ln|tan(arctan(15 x))+sec(arctan(15 x))|)

25(sec²(arctan(15 x))sin(arctan(15 x))2 +12 ln|tan(arctan(15 x))+sec(arctan(15 x))|) بسّط: 25(150 x√25+x²+12 ln(|x+√25+x²|5 ))

=25(150 x√25+x²+12 ln(|x+√25+x²|5 ))

أضف ثابت التكامل للحل

=25(150 x√25+x²+12 ln(|x+√25+x²|5 ))+C

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

vertices (-3x^2-2x+6)/5

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