12 Days of Christmas Integrals: ∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx
HAPPY NEW YEAR!!!! HERE'S TO A WONDERFUL 2022! May we make our future and present bright! (We all could use a bright year, given how the last few years went)
On the Eighth day of Christmas Integrals, the integral featured today is...
∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx
By division:
= ∫ 1 + (2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) dx
Find the partial fraction of (2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8)
(2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) = A ÷ (x - 2) + B ÷ (x - 4)
2 ∙ x - 5 = A ∙ (x - 4) + B ∙ (x - 2)
2 ∙ x - 5 = A ∙ x - 4 ∙ A + B ∙ x - 2 ∙ B
which implies that:
-5 = -4 ∙ A - 2 ∙ B
2 = A + B
Solving for A and B:
A = 1/2 and B = 3/2
(2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) = (1/2) ÷ (x - 2) + (3/2) ÷ (x - 4)
= ∫ 1 + (1/2) ÷ (x - 2) + (3/2) ÷ (x - 4) dx
= x + 1/2 ∙ ln|x - 2| + 3/2 ∙ ln|x - 4| + C
Eddie
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