Begin by splitting the fraction into partial fractions:
#(x-2)/((x+2)(x-3)) = A/(x+2) + B/(x-3)#
Now multiply by #(x+2)(x+3)# to find #A# and #B#.
# x-2 = (x-3)A + (x+2)B#
Set #x=3# to find #B#.
#(3)-2 = ((3)-3)A+ ((3)+2)B = 5B -> 5B =1#
# therefore B =1/5#
Set #x = -2# to find #A# so we will have:
#-4 =-5A therefore A = 4/5#
So we can now rewrite the integral in partial fractions and integrate:
#int (x-2)/((x+2)(x-3))dx = int 4/(5(x+2)) + 1/(5(x-3))dx#
# therefore f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) + C#
Now you can use the condition specified: #f(2) = -1#
#-> -1 = 4/5lnAbs(4) -1/5lnAbs(-1) + C#
#lnAbs(-1) = ln(1) = 0# which leaves us with:
#C = -1-4/5ln(4)# which means our final answer is:
#f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) -1-4/5ln(4)#
