Recall that x^4 - 2x^2 - 8 can be factored as (x^2 - 4)(x^2 + 2) = (x + 2)(x - 2)(x^2 + 2).
Hence:
A/(x + 2) + B/(x - 2) + (Cx + D)/(x^2 + 2) = x^2/((x + 2)(x - 2)(x^2 + 2))
A(x - 2)(x^2 + 2) + B(x + 2)(x^2 + 2) + (Cx + D)(x^2 - 4) = x^2
A(x^3 - 2x^2 + 2x - 4) + B(x^3 + 2x^2 + 2x + 4) + Cx^3 + Dx^2 - 4Cx - 4D = x^2
Ax^3 - 2Ax^2 + 2Ax - 4A + Bx^3 + 2Bx^2 + 2Bx + 4B + Cx^3 + Dx^2 - 4Cx - 4D = x^2
Write a system of equations.
{(A+ B + C = 0), (2B - 2A + D = 1), (2A + 2B - 4C = 0), (4B - 4A - 4D = 0):}
After solving, either using a computer algebra system or tedious algebra, you should get that A = -1/6, B = 1/6, C = 0 and D = 1/3.
=int-1/(6(x + 2)) + 1/(6(x - 2)) + 1/(3(x^2 + 2))dx
The first two terms can be integrated using the rule int1/xdx = ln|x| + C. Therefore, we are left with:
=1/6ln|x- 2| - 1/6ln|x + 2| + int1/(3(x^2 + 2))dx
We need to recognize the last integral as one integrable by arctangent. Use the substitution u = x/sqrt(2).
int1/(3(x^2 + 2))dx = 1/(3sqrt(2))int1/(u^2 + 1)
The integral of 1/(x^2 + 1)dx = arctanx:
int1/(3(x^2 + 2))dx = arctanu/(3sqrt(2))
int1/(3(x^2 + 2))dx = arctan(x/sqrt(2))/(3sqrt(2)) + C
Put this together
=1/6ln|x- 2| - 1/6ln|x + 2| + arctan(x/sqrt(2))/(3sqrt(2)) +C
Hopefully this helps!
