Q.127. Integration

Question 127.
Evaluate ∫ (3x^2 + 3x - 56) / (x^3 - 4x^2 + 22x + 68) dx.

Answer 127.
x^3 - 4x^2 + 22x + 68

= x^3 + 2x^2 - 6x^2 - 12x + 34x + 68
= x^2 (x + 2) - 6x (x + 2) + 34 (x + 2)
= (x + 2) (x^2 - 6x + 34)

Let
(3x^2 + 3x - 56) / (x^3 - 4x^2 + 22x + 68)
= (3x^2 + 3x - 56) / (x + 2) (x^2 - 6x + 34)
= A/(x + 2) + (Bx + C)/(x^2 - 6x + 34)

=> 3x^2 + 3x - 56 = A (x^2 - 6x + 34) + (Bx + C) (x + 2)
x = - 2 => A = - 50/50 = - 1
Comparing coefficients of x^2,
A + B = 3 => B = 4
Comparing constant terms,
34A + 2C = - 56 => C = - 11

=> ∫ (3x^2 + 3x - 56) / (x^3 - 4x^2 + 22x + 68) dx
= - ∫ dx /(x + 2) + ∫ (4x - 11) / (x^2 - 6x + 34) dx
= - ln l x + 2 l + 2 ∫ (2x - 6 + 1/2) / (x^2 - 6x + 34) dx
= - ln l x + 2 l + 2 ∫ d(x^2 - 6x + 34) / (x^2 - 6x + 34) + ∫ dx / [(x - 3)^2 + (5)^2]
= - ln l x + 2 l + 2 ln (x^2 - 6x + 34) + (1/5) arctan [(x - 3)/5] + c.

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