#int 3/((x-2)(x+1))dx#
The method of partial fractions involves splitting the denominator into its factors. Lets ignore the integral for now. We are looking for two values, #A# and #B# such that;
#A/(x-2) + B/(x+1) = 3/((x-2)(x+1))#
Now we need to combine #A# and #B# over a common denominator.
#A/((x-2))((x+1))/((x+1)) + B/((x+1)) ((x-2))/((x-2)) = 3/((x-2)(x+1))#
#(A(x+1)+B(x-2))/((x-2)(x+1)) = 3/((x-2)(x+1))#
We can cancel the denominator on both sides, leaving;
#A(x+1)+B(x-2) = 3#
Now we can solve for #A# and #B# by selectively choosing our values of #x# so that the terms in parenthesis are #0#. Let #x=-1#.
#Acolor(red)cancel(color(black)((-1+1)))^0 + B(-1-2) = 3#
#-3B = 3#
#B=-1#
Let #x=2#.
#A(2+1)+Bcolor(red)cancel(color(black)((2-2)))^0 = 3#
#3A = 3#
#A=1#
Now that we have our values for #A# and #B# we can split apart our integral by plugging them in.
#int 3/((x-2)(x+1)) dx = int (1/(x-2) +(- 1)/(x+1)) dx#
#=int 1/(x-2) dx - int 1/(x+1) dx#
Now we have two integrals that we can solve using substitution. Let #u_1=x-2# for the first integral and #u_2=x+1# for the second.
#int 1/u_1 du_1 - int 1/u_2 du_2#
#ln(u_1) - ln(u_2) +C#
#ln((u_1)/(u_2)) + C#
Plug in the values for #u_1# and #u_2#.
#=ln((x-2)/(x+1)) +C#
