#int (3x-1)/(x^2+2x-8)dx=int (3x-1)/((x+4)(x-2))dx#
To decompose the function into partial fractions, there are #A# and #B# such that:
#A/(x+4)+B/(x-2)=(3x-1)/((x+4)(x-2))#
#A(x-2)+B(x+4)=3x-1#
#(A+B)x+(4B-2A)=3x-1#
Now solve for #A# and #B#.
#A+B=3#
#4B-2A=-1#
#->A=13/6, B=5/6#
Integrate:
#int (13/6)/(x+4)dx+int (5/6)/(x-2)dx#
#13/6 int1/(x+4)dx+5/6 int 1/(x-2)dx#
Set #u=x+4# in the first and #v=x-2# in the second so that both have derivative of #1#.
#13/6 int1/u du+5/6int1/v dv#
#13/6 lnu+5/6 ln v#
#13/6 ln abs(x+4)+ 5/6 ln abs(x-2)#
#(13ln abs(x+4)+ 5ln abs(x-2))/6#
