Start from the given integral and set up the variables using A, B, C ,D, E
#int (x^3)/((x^4-16)(x-3))dx=#
#int((Ax+B)/(x^2+4)+C/(x+2)+D/(x-2)+E/(x-3))dx#
and out of the numerators from both sides we have
#x^3=Ax^4-4Ax^2-3Ax^3+12Ax+Bx^3-4Bx-3Bx^2+12B+Cx^4-5Cx^3+10Cx^2-20Cx+24C+Dx^4-Dx^3-2Dx^2-4Dx-24D+Ex^4-16E#
After setting up all the equations . The following are the equations
#C+D+E=0" "#equation 1
#B-5C-D=1" "#equation 2
#-3B+10C-2D=0" "#equation 3
#-4B-20C-4D=0" "#equation 4
#12B+24C-24D-16E=0" "#equation 5
Solving for A, B, C, D, E by any method
#A=(-3)/26#
#B=2/13#
#C=(-1)/20#
#D=(-1)/4#
#E=27/65#
We have to integrate now
#int (x^3)/((x^4-16)(x-3))dx=#
#int(((-3)/26x+2/13)/(x^2+4)+(-1)/20/(x+2)+(-1)/4/(x-2)+27/65/(x-3))dx#
#int (x^3)/((x^4-16)(x-3))=(-3)/52ln(x^2+4)+1/13tan^-1 (x/2)-1/20ln(x+2)-1/4ln(x-2)+27/65ln(x-3)+C_0#
where #C_0# be the constant of integration
God bless....I hope the explanation is useful.
