Question

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

Answer 1

`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.