Write the partial fractions equation:
#(7x-2)/((x-3)^2(x+1)) = A/(x+1)+B/(x-3)+C/(x-3)^2#
Multiply both sides by #(x-3)^2(x+1)#:
#7x-2 = A(x-3)^2+B(x-3)(x+1)+C(x+1)#
Let #x = -1#
#7(-1)-2 = A(-1-3)^2+B(-1-3)(-1+1)+C(-1+1)#
#-9 = 16A#
#A = -9/16#
#7x-2 = -9/16(x-3)^2+B(x-3)(x+1)+C(x+1)#
Let #x = 3#
#7(3)-2 = -9/16(3-3)^2+B(3-3)(3+1)+C(3+1)#
#C = 19/4
#7x-2 = -9/16(x-3)^2+B(x-3)(x+1)+19/4(x+1)#
Let #x = 0#
#7(0)-2 = -9/16(0-3)^2+B(0-3)(0+1)+19/4(0+1)#
#-2 = -81/16-3B+19/4(0+1)#
#B = 9/16#
The partial fractions are:
#(7x-2)/((x-3)^2(x+1)) = -9/16 1/(x+1)+9/16 1/(x-3)+19/4 1/(x-3)^2#
Integrate:
#int (7x-2)/((x-3)^2(x+1)) dx = -9/16 int 1/(x+1) dx+9/16 int 1/(x-3) dx +19/4 int 1/(x-3)^2 dx#
The integrals are trivial:
#int (7x-2)/((x-3)^2(x+1)) dx = -9/16 lnt|x+1|+9/16 ln|x-3| -19/4 1/(x-3)+ C#
