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integral of (x^2-6x+8)/(x-4)

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∫ x²−6x+8x−4 dx

Solution

2x+12 (x−4)²−8+C

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∫ x²−6x+8x−4 dx

Expand x²−6x+8x−4 : x²x−4 −6xx−4 +8x−4

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

=∫ x²x−4 dx−∫ 6xx−4 dx+∫ 8x−4 dx

∫ x²x−4 dx=(x−4)²2 +8(x−4)+16ln|x−4|

∫ 6xx−4 dx=6(x−4+4ln|x−4|)

∫ 8x−4 dx=8ln|x−4|

=(x−4)²2 +8(x−4)+16ln|x−4|−6(x−4+4ln|x−4|)+8ln|x−4|

Simplify (x−4)²2 +8(x−4)+16ln|x−4|−6(x−4+4ln|x−4|)+8ln|x−4|: 2x+12 (x−4)²−8

=2x+12 (x−4)²−8

Add a constant to the solution

=2x+12 (x−4)²−8+C

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−▭▭	<	7	8	9	÷	`AC`
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integral of (x^2-6x+8)/(x-4)

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