integral of (1-cos(2x))/x

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∫ √x−3x−9 dx

Solution

−6(12 ln|√x3 +1|−12 ln|√x3 −1|−13 √x)−3ln|x−9|+C

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∫ √x−3x−9 dx

Expand √x−3x−9 : √xx−9 −3x−9

Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ √xx−9 dx−∫ 3x−9 dx

∫ √xx−9 dx=−6(12 ln|√x3 +1|−12 ln|√x3 −1|−13 √x)

∫ 3x−9 dx=3ln|x−9|

=−6(12 ln|√x3 +1|−12 ln|√x3 −1|−13 √x)−3ln|x−9|

Add a constant to the solution

=−6(12 ln|√x3 +1|−12 ln|√x3 −1|−13 √x)−3ln|x−9|+C

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

integral of (1-cos(2x))/x

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