#x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x + 2)(x - 2)(x^2 + 4)#
The partial fraction decomposition will therefore be of the form #A/(x + 2) + B/(x - 2) + (Cx + D)/(x^2 + 4)#.
#A/(x + 2)+ B/(x- 2) + (Cx + D)/(x^2 + 4) = 1/((x + 2)(x - 2)(x^2 + 4))#
#A(x - 2)(x^2 + 4) + B(x + 2)(x^2 + 4) + (Cx + D)(x +2)(x - 2) = 1#
#A(x^3 - 2x^2 + 4x - 8) + B(x^3 + 2x^2 + 4x + 8) + (Cx + D)(x^2 - 4)= 1#
#Ax^3- 2Ax^2 + 4Ax - 8A + Bx^3 + 2Bx^2 + 4Bx + 8B + Cx^3 + Dx^2 - 4Cx - 4D = 1#
#(A + B + C)x^3 + (2B - 2A + D)x^2 + (4A + 4B - 4C)x + (8B - 8A - 4D) = 1#
Write a system of equations now:
#{(A + B + C = 0),(2B - 2A + D = 0), (4A + 4B - 4C = 0), (8B - 8A - 4D = 1
):}#
First of all, the third equation can be simplified to #A + B - C = 0#. We can get rid of the #C# variable by adding the first and the third equations, to get #2A + 2B = 0 -> A + B = 0#. This means that #B = -A#. Also, from the second equation, we deduce that #D = 2A - 2B#.
Substitute into the fourth equation:
#8(-A) - 8A - 4(2A - 2(-A)) = 1#
#-8A - 8A - 8A - 8A = 1#
#-32A = 1#
#A = -1/32#
Solving for the other variables is now relatively simple. You should get #A = -1/32#, #B = 1/32#,#C = 0# and #D = -1/8# as final answers.
#:.#The partial fraction decomposition is #-1/(32(x + 2)) + 1/(32(x - 2)) - 1/(8(x^2 + 4))#
Let's look closely at the integration of #int-1/(8(x^2 + 4))dx#. This will be of the arctangent form, but first we need to perform a u-substitution.
Let #u = x/2#. Then #du = 1/2dx#.
#-1/8int1/(u^2 +1) * 1/2du#
#-1/16int 1/(u^2 + 1)du#
This is a standard integral.
#-1/16arctanu#
#-1/16arctan(x/2)#
Now deal with the other parts of the problem. The other two terms can be integrated using #int1/xdx = ln|x| + C#.
#int1/(32(x - 2))dx = 1/32ln|x - 2|#
#int-1/(32(x +2))dx = -1/32ln|x + 2|#
Therefore, #int1/(x^4 - 16)dx = 1/32ln|x - 2| - 1/32ln|x + 2| - 1/16arctan(x/2) + C#
Hopefully this helps!
