Saturday, January 1, 2022

12 Days of Christmas Integrals: ∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

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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