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

Feb 20, 2017

$\frac{1}{4} \left(6 x + 5 \ln | 2 x - 3 |\right) + C$

#### 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.

$f \left(x\right) = \frac{3 x - 2}{2 x - 3} = \left(3 x - 2\right) {\left(2 x - 3\right)}^{-} 1$

We are trying to find:

$\int \left(3 x - 2\right) {\left(2 x - 3\right)}^{-} 1 \mathrm{dx}$

Let $u = 2 x - 3$

$\frac{\mathrm{du}}{\mathrm{dx}} = 2$

$\mathrm{dx} = \frac{1}{2} \mathrm{du}$

$x = \frac{u + 3}{2}$

$3 \left(\frac{u + 3}{2}\right) - 2 = \frac{3 u + 9}{2} - \frac{4}{2} = \frac{3 u + 5}{2}$

$\int \left(3 x - 2\right) {\left(2 x - 3\right)}^{-} 1 \mathrm{dx} = \int \frac{3 u + 5}{2} \left({u}^{-} 1\right) \frac{1}{2} \mathrm{du} =$

$\frac{1}{4} \int \left(3 u + 5\right) \left({u}^{-} 1\right) \mathrm{du} = \frac{1}{4} \int 3 + 5 {u}^{-} 1 \mathrm{du}$

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).

$\frac{3}{4} \int 1 \mathrm{du} + \frac{5}{4} \int {u}^{-} 1 \mathrm{du} =$

We need to remember that we can't integrate ${u}^{-} 1$ using the power rule, but we can use the identity $\int {u}^{-} 1 = \ln | u |$.

$\frac{3}{4} u - \frac{5}{4} \ln | u | + C$

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

$\frac{3}{4} \left(2 x - 3\right) + \frac{5}{4} \ln | 2 x - 3 | + C =$

$\frac{1}{4} \left(6 x - 9 + 5 \ln | 2 x - 3 |\right) + C$

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