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
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.
To find the indefinite integral of ∫ (x^2 - 8)/ (x^2 - 16) dx, you followed the correct process by splitting the fraction into two separate integrals. Let's go through the steps again and see why you didn't end up with the same solution.
You correctly calculated the integral of (x^2 - 8) as [(x^3)/3 - 8x] + C.
Now, let's focus on the integral of 1/(x^2 - 16). To integrate this, you used partial fraction decomposition by breaking it down into two fractions with different denominators.
∫ (1/(x^2 - 16)) dx = ∫ [(-1/8)/(x + 4)] + [(1/8)/(x - 4)] dx
Here, you made a small error when integrating the terms. The correct integrals for the two fractions should be:
∫ (-1/8)/(x + 4) dx = (-1/8)*ln|x + 4| + C1
∫ (1/8)/(x - 4) dx = (1/8)*ln|x - 4| + C2
Notice that the natural logarithm function has an absolute value around the argument. This is because the logarithm is only defined for positive values, and the absolute value ensures that negative values also work.
So, when combining the two integrals, the correct expression would be:
∫ (x^2 - 8)/ (x^2 - 16) dx = [(x^3)/3 - 8x] * (-1/8)*ln|x + 4| + (1/8)*ln|x - 4| + C
Now, comparing this with the given solution of x + ln(|x - 4| / |x + 4|) + C, we can see that they have canceled out the common factor of (-1/8) and also simplified the term (x^3)/3. This simplification was likely done separately and then factored out of the integral to make the solution more concise.
In summary, your integral expression is correct, but the given solution has further simplified the terms involving -8x and (x^3)/3.