range of 2sin(5x)

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

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

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∫ √x2x+3 dx

Solution

√x−√32 arctan(√23 √x)+C

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∫ √x2x+3 dx

فعّل طريقة التكامل بالتعويض: ∫ 2u²2u²+3 du

=∫ 2u²2u²+3 du

∫a·f(x)dx=a·∫f(x)dx :قم بإخراج الثابت

=2· ∫ u²2u²+3 du

u²2u²+3 استخدم القسمة الطويلة لـ: 12 −32(2u²+3)

=2· ∫ 12 −32(2u²+3) du

∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx :فعّل الميزة الخطيّة

=2(∫ 12 du−∫ 32(2u²+3) du)

∫ 12 du=12 u

∫ 32(2u²+3) du=√32√2 arctan(√23 u)

=2(12 u−√32√2 arctan(√23 u))

u=√x استبدل مجددًا

=2(12 √x−√32√2 arctan(√23 √x))

2(12 √x−√32√2 arctan(√23 √x)) بسّط: √x−√32 arctan(√23 √x)

=√x−√32 arctan(√23 √x)

أضف ثابت التكامل للحل

=√x−√32 arctan(√23 √x)+C

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

range of 2sin(5x)

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