midpoint (a, b), (c, d)

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

☐²

x^☐

√☐

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log_□

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ddx

≥	≤	·	÷	`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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range 2cos(x)

Solution

−2≤ f(x)≤ 2

Interval Notation

[−2, 2]

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Solution steps

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∫ xln(x²+3)dx

Apply u-substitution: ∫ ln(u)2 du

=∫ ln(u)2 du

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

=12 · ∫ ln(u)du

Apply Integration By Parts: uln(u)−∫ 1du

=12 (uln(u)−∫ 1du)

∫ 1du=u

=12 (uln(u)−u)

Substitute back u=x²+3

=12 ((x²+3)ln(x²+3)−(x²+3))

Expand (x²+3)ln(x²+3)−(x²+3): x²ln(x²+3)+3ln(x²+3)−x²−3

=12 (x²ln(x²+3)+3ln(x²+3)−x²−3)

Add a constant to the solution

=12 (x²ln(x²+3)+3ln(x²+3)−x²−3)+C

Solve instead

domain 2cos(x)

periodicity 2cos(x)

amplitude 2cos(x)

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

midpoint (a, b), (c, d)

Solution

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