# How do you integrate f(x)=(3x-4)/(2x+1) using the quotient rule?

Dec 20, 2016

$\int f \left(x\right) \mathrm{dx} = \frac{3}{2} x - \frac{11}{4} \ln | 2 x + 1 | + C$

#### Explanation:

There is no such thing as a quotient rule for integration!

$f \left(x\right) = \frac{3 x - 4}{2 x + 1}$

To integrate this particular function we can use the following substitution:

$\left.\begin{matrix}\text{Let " & u=2x+1 & => x=1/2u-1/2 & => 3x-4=3/2u-11/2 \\ "Then } & \frac{\mathrm{du}}{\mathrm{dx}} = 2 & \implies \mathrm{dx} = \frac{1}{2} \mathrm{du} & \null\end{matrix}\right.$

And so:

$\int \frac{3 x - 4}{2 x + 1} \mathrm{dx} = \int \frac{\frac{3}{2} u - \frac{11}{2}}{u} \frac{1}{2} \mathrm{du}$
$\text{ } = \frac{1}{4} \int \frac{3 u - 11}{u} \mathrm{du}$
$\text{ } = \frac{1}{4} \int 3 - \frac{11}{u} \mathrm{du}$
$\text{ } = \frac{1}{4} \left\{3 u - 11 \ln | u |\right\} + C '$
$\text{ } = \frac{1}{4} \left\{3 \left(2 x + 1\right) - 11 \ln | 2 x + 1 |\right\} + C '$
$\text{ } = \frac{1}{4} \left\{6 x + 3 - 11 \ln | 2 x + 1 |\right\} + C '$
 " " = 3/2x+3/4-11/4ln|2x+1|} + C'
$\text{ } = \frac{3}{2} x - \frac{11}{4} \ln | 2 x + 1 | + C$