integral of (4x^2+x+9)/(4x^2-9)

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∫ 4x²+x+94x²−9 dx

Solution

x−34 ln(13 |2x+3|)+34 ln(13 |2x−3|)+18 ln|4x²−9|−34 ln|23 x+1|+34 ln|23 x−1|+C

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∫ 4x²+x+94x²−9 dx

Expand 4x²+x+94x²−9 : 4x²4x²−9 +x4x²−9 +94x²−9

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

=∫ 4x²4x²−9 dx+∫ x4x²−9 dx+∫ 94x²−9 dx

∫ 4x²4x²−9 dx=−14 (−4x+3ln(|2x+3|3 )−3ln(|2x−3|3 ))

∫ x4x²−9 dx=18 ln|4x²−9|

∫ 94x²−9 dx=−34 (ln|23 x+1|−ln|23 x−1|)

=−14 (−4x+3ln(|2x+3|3 )−3ln(|2x−3|3 ))+18 ln|4x²−9|−34 (ln|23 x+1|−ln|23 x−1|)

Simplify −14 (−4x+3ln(|2x+3|3 )−3ln(|2x−3|3 ))+18 ln|4x²−9|−34 (ln|23 x+1|−ln|23 x−1|): x−34 ln(13 |2x+3|)+34 ln(13 |2x−3|)+18 ln|4x²−9|−34 ln|23 x+1|+34 ln|23 x−1|

=x−34 ln(13 |2x+3|)+34 ln(13 |2x−3|)+18 ln|4x²−9|−34 ln|23 x+1|+34 ln|23 x−1|

Add a constant to the solution

=x−34 ln(13 |2x+3|)+34 ln(13 |2x−3|)+18 ln|4x²−9|−34 ln|23 x+1|+34 ln|23 x−1|+C

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

integral of (4x^2+x+9)/(4x^2-9)

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