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!
