integral of (4x^2-2x+1)/(2x+3)

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∫ 4x²−2x+12x+3 dx

Solution

x²−4x+132 ln|2x+3|−32 −274 +C

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∫ 4x²−2x+12x+3 dx

Expand 4x²−2x+12x+3 : 4x²2x+3 −2x2x+3 +12x+3

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

=∫ 4x²2x+3 dx−∫ 2x2x+3 dx+∫ 12x+3 dx

∫ 4x²2x+3 dx=14 (4x²−12x+18ln|2x+3|−27)

∫ 2x2x+3 dx=x+32 −32 ln|2x+3|

∫ 12x+3 dx=12 ln|2x+3|

=14 (4x²−12x+18ln|2x+3|−27)−(x+32 −32 ln|2x+3|)+12 ln|2x+3|

Simplify 14 (4x²−12x+18ln|2x+3|−27)−(x+32 −32 ln|2x+3|)+12 ln|2x+3|: x²−4x+132 ln|2x+3|−32 −274

=x²−4x+132 ln|2x+3|−32 −274

Add a constant to the solution

=x²−4x+132 ln|2x+3|−32 −274 +C

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
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integral of (4x^2-2x+1)/(2x+3)

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