Posted by **Maryam** on Sunday, May 26, 2013 at 3:33am.

1. Evaluate the following integrals

(a) 4x2 +6x−12 / x3 − 4x dx

- Calculus -
**Reiny**, Sunday, May 26, 2013 at 7:48am
I will assume you meant

(4x^2 +6x−12)/(x^3 − 4x)

= (4x^2 + 6x - 12)/( x(x+2)(x-2) )

split it into partial fractions

let

A/x + B/(x+2) + C/(x-2) = (4x^2 + 6x - 12)/( x(x+2)(x-2) )

multiply by x(x+2)x-2)

A(x^2-4) + Bx(x-2) + Cx(x+2) = 4x^2 + 6x - 12

This must be true for all x's

let x = 0

-4A = -12

A=3

let x = 2

8C = 16 + 12-12 = 16

C = 2

let x = -2

8B = 16 + 12-12 = -8

B = -1

so ∫ (4x^2 + 6x - 12)/( x(x+2)(x-2) ) dx

= ∫3/x dx + ∫-1/(x+2)dx + ∫2/(x-2) dx

= 3ln x - ln(x+2) + 2ln(x-2) + k, where k is a constant

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