# Calculus

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1. Evaluate the following integrals
(a) 􏰀 4x2 +6x−12 / x3 − 4x dx

• Calculus -

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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