Question

Find the following indefinite integral
∫ (x^2 -8)/ (x^2 -16) dx

My work
∫ (x^2 - 8) * ∫ (1/(x^2 - 16)

∫ (x^2 - 8) = [(x^3)/(3) - 8x)] + C

∫ (1/(x^2 - 16) = ∫ [(-1/8)/(x+4)] + [(1/8)/(x-4)] = (-1/8)*ln|(x+4)| +(1/8)*|(x-4)| + C

So I get this,
∫ (x^2 -8)/ (x^2 -16) dx = [(x^3)/(3) - 8x)] * (-1/8)*ln|(x+4)| +(1/8)*ln|(x-4)|

The Solution given is = x + ln ( |(x-4)| / |(x+4)| ) + C

I can see that they probably remove the 1/8 by cancelling with the -8 from -8x, but I do not ee how they got rid of (x^3)/(3)

Thx

Answer 1

NO, you cannot just take the integral of each of the factors, and them combine them like you did.

first of all, I did a long division:
(x^2 - 8)/(x^2 - 16) = 1 + 8/(x^2 - 16 = 1 + 8/((x-4)(x+4))

So let's find the partial fractions for 8/((x-4)(x+4))
let 8/((x-4)(x+4)) = A/(x-4) + B(x+4)
= (A(x+4) + B(x-4))/(((x-4)(x+4))

then A(x+4) + B(x-4) = 8
let x=8, 8A + 0 = 8 , A = 1
let x = -8, 0 -8B = 8 , B = -1

Sooo....
(x^2 -8)/ (x^2 -16) = 1 + 1/(x-4) - 1/(x+4)
∫ (x^2 -8)/ (x^2 -16) dx = ∫ (1 + 1/(x-4) - 1/(x+4) ) dx
= x + ln |x-4| - ln|x + 4| + c
or
x + ln( |x-4| / |x + 4| ) + c , using log rules , which is their answer.