Question

How do you integrate `f(x)=(3x-2)/(2x-3)` using the quotient rule?

Answer 1

`1/4(6x+5ln|2x-3|)+C`

Explanation:

This answer will assume the questioner is mistaken about the quotient rule.

There is no quotient rule for integration; this needs to be integrated via substitution or another apt method.

`f(x)=(3x-2)/(2x-3)=(3x-2)(2x-3)^-1`

We are trying to find:

`int(3x-2)(2x-3)^-1dx`

Let `u=2x-3`

`(du)/dx=2`

`dx=1/2du`

`x=(u+3)/2`

`3((u+3)/2)-2=(3u+9)/2-4/2=(3u+5)/2`

`int(3x-2)(2x-3)^-1dx=int(3u+5)/2(u^-1)1/2du=`

`1/4int(3u+5)(u^-1)du=1/4int3+5u^-1 du`

We can now separate the integral into two parts. (It is a good idea to take out any constants so that the integral is as simple as possible).

`3/4int1du+5/4intu^-1du=`

We need to remember that we can't integrate `u^-1` using the power rule, but we can use the identity `intu^-1=ln|u|`.

`3/4u-5/4ln|u|+C`

Now substitute `u` for `2x-3`

`3/4(2x-3)+5/4ln|2x-3|+C=`

`1/4(6x-9+5ln|2x-3|)+C`

Since `9` is a constant, we can 'absorb' it into the constant of integration.