How do you integrate #int ( 1-3x)/((2x-1)(x+2))# using partial fractions?

1 Answer

#color(blue)(int (1-3x)/((2x-1)(x+2))dx= -1/10*ln (2x-1)-7/5*ln(x +2) +C)#

Explanation:

From the given #int (1-3x)/((2x-1)(x+2))dx#
we use partial fraction with the variables #A# and #B#
#int (1-3x)/((2x-1)(x+2))dx=int A/(2x-1)dx+B/(x +2) dx#

Let us determine the constants first
#(1-3x)/((2x-1)(x+2))= A/(2x-1)+B/(x +2) #
#(1-3x)/((2x-1)(x+2))= (A(x+2)+B(2x-1))/((2x-1)(x +2)) #
#(1-3x)/((2x-1)(x+2))= (Ax+2A+2Bx-B)/((2x-1)(x +2)) #
Now we can set up the equations

#A+2B=-3#
#2A-B=1#

Simultaneous solution results to:
#A=-1/5#
#B=-7/5#

Now, we can integrate
#int (1-3x)/((2x-1)(x+2))dx=int A/(2x-1)dx+B/(x +2) dx#
#int (1-3x)/((2x-1)(x+2))dx=int (-1/5)/(2x-1)dx+(-7/5)/(x +2) dx#
#int (1-3x)/((2x-1)(x+2))dx= -1/10*ln (2x-1)-7/5*ln(x +2) +C#

God bless ..... I hope the explanation is useful.