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integral of (sqrt(x))/(2x+3)

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∫ √x2x+3 dx

Solution

√x−√32 arctan(√23 √x)+C

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∫ √x2x+3 dx

Apply u-substitution: ∫ 2u²2u²+3 du

=∫ 2u²2u²+3 du

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

=2· ∫ u²2u²+3 du

Long division u²2u²+3 : 12 −32(2u²+3)

=2· ∫ 12 −32(2u²+3) du

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

=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))

Substitute back u=√x

=2(12 √x−√32√2 arctan(√23 √x))

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

=√x−√32 arctan(√23 √x)

Add a constant to the solution

=√x−√32 arctan(√23 √x)+C

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

integral of (sqrt(x))/(2x+3)

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