領域 2cos(x)

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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

simplify solve for inverse tangent line

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∫ 2x²+8x+17 dx

Solution

2arctan(x+4)+C

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∫ 2x²+8x+17 dx

Take the constant out: ∫a·f(x)dx=a·∫f(x)dx

=2· ∫ 1x²+8x+17 dx

Complete the square x²+8x+17: (x+4)²+1

=2· ∫ 1(x+4)²+1 dx

Apply u-substitution: ∫ 1u²+1 du

=2· ∫ 1u²+1 du

Use the common integral: ∫ 1u²+1 du=arctan(u)

=2arctan(u)

Substitute back u=x+4

=2arctan(x+4)

Add a constant to the solution

=2arctan(x+4)+C

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

領域 2cos(x)

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