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

1 Answer
Feb 20, 2017

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.