Firstly, you need the #dx# at the end of the integral for formal presentation i.e. #int (x^3 + 2)/(x^4-16) dx#.
Secondly, you need to decompose #(x^3 + 2)/(x^4-16)# into partial fractions. This is made simpler by noting that
#x^4-16 = (x^2-4)(x^2+4) = (x+2)(x-2)(x^2+4)#
That is,
#(x^3 + 2)/(x^4-16) = A/(x+2) + B/(x-2) + (Cx+D)/(x^2+4)#
for real constants #A#, #B#, #C# and #D#.
Using the cover-up rule and comparison of coefficients by algebraic means, you should be able to find solve for the unknowns, that is, #A = 3/16#, #B = 5/16#, #C=1/2# and #D=–1/4#.
Thus, #(x^3 + 2)/(x^4-16) = (3/16)/(x+2) + (5/16)/(x-2) + (1/2x-1/4)/(x^2+4)#, and
#int (x^3 + 2)/(x^4-16) dx=int ((3/16)/(x+2) + (5/16)/(x-2) + (1/2x-1/4)/(x^2+4)) dx#.
The first two integrals are simple, i.e.
#int (3/16)/(x+2) dx = 3/16 int 1/(x+2) dx = 3/16 ln abs(x+2)+K_1#
and likewise,
#int (5/16)/(x+2) dx = 5/16 ln abs(x-2)+K_2#
The third part is a little tricky. By appropriate decompositions,
#int (1/2x-1/4)/(x^2+4) dx = 1/4 int (2x)/(x^2+4) dx - 1/4 int 1/(x^2+4) dx
= 1/4 ln(x^2+4) - 1/4(1/2)arctan (x/2) + K_3 = 1/4 ln(x^2+4) - 1/8 arctan (x/2) + K_3#
Thus, putting everything together,
#int (x^3 + 2)/(x^4-16) dx=3/16 ln abs(x+2) + 5/16 ln abs(x-2) + 1/4 ln(x^2+4) - 1/8 arctan (x/2) + K#
where #K = K_1 + K_2 + K_3#.
