How do you integrate $[((3x^2)+3x+12) / ((x-5)(x^2+9))]dx$ using partial fractions?

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Answer 1 · 2016-11-12T20:39:38

Please see the explanation

Expand the fraction:

$(3x^2 + 3x + 12)/((x - 5)(x^2 + 9)) = A/(x - 5) + (Bx + C)/(x^2 + 9)$

Multiply both sides by the left side denominator:

$3x^2 + 3x + 12 = A(x^2 + 9) + (Bx + C)(x - 5)$

Let x = 5 to make B and C disapper

$3(5)^2 + 3(5) + 12 = A(5^2 + 9)$

$102 = A(34)$

$A = 3$

Substitute 3 for A:

$3x^2 + 3x + 12 = 3(x^2 + 9) + (Bx + C)(x - 5)$

Let x = 0 to make B disappear:

$12 = 3(9) + C(-5)$

$-15 = -5C$

$C = 3$

Substitute 3 for C:

$3x^2 + 3x + 12 = 3(x^2 + 9) + (Bx + 3)(x - 5)$

Let x = 1:

$3 + 3 + 12 = 3(1 + 9) + (B + 3)(1 - 5)$

$18 = 30 -4B - 12$

$B = 0$

$int((3x^2 + 3x + 12)/((x - 5)(x^2 + 9)))dx = 3int1/(x - 5)dx + 3int1/(x^2 + 5)dx$

$int((3x^2 + 3x + 12)/((x - 5)(x^2 + 9)))dx = 3ln|x - 5| + 3sqrt(5)/5tan^-1((sqrt(5))x /5) + C$

How do you integrate [((3x^2)+3x+12) / ((x-5)(x^2+9))]dx[((3x^2)+3x+12) / ((x-5)(x^2+9))]dx using partial fractions?