# How do you integrate f(x)=log_6(2x)/lnx using the quotient rule?

Nov 30, 2017

$\frac{\ln \left(2\right) l i \left(x\right) + x}{\ln} \left(6\right) + C$

#### Explanation:

The quotient rule is only for derivatives, it can't be applied to integrals.

To solve the integral, we're first going to use the following rule:
${\log}_{a} \left(b\right) = {\log}_{x} \frac{b}{\log} _ x \left(a\right)$ where ${\log}_{x}$ can be any logarithm.

We'll simplify to the natural logarithm, since we have that on the bottom:
$\int \setminus {\log}_{6} \frac{2 x}{\ln} \left(x\right) \setminus \mathrm{dx} = \int \setminus \frac{\ln \frac{2 x}{\ln} \left(6\right)}{\ln} \left(x\right) \setminus \mathrm{dx} = \int \setminus \ln \frac{2 x}{\ln} \left(6\right) \cdot \frac{1}{\ln} \left(x\right) \setminus \mathrm{dx}$

Now we can take the constant out:
$\frac{1}{\ln} \left(6\right) \int \setminus \ln \frac{2 x}{\ln} \left(x\right) \setminus \mathrm{dx}$

We can also use the rule that ${\log}_{x} \left(a b\right) = {\log}_{x} \left(a\right) + {\log}_{x} \left(b\right)$ to rewrite in the following way:
$\frac{1}{\ln} \left(6\right) \int \setminus \frac{\ln \left(2\right) + \ln \left(x\right)}{\ln} \left(x\right) \setminus \mathrm{dx} = \frac{1}{\ln} \left(6\right) \int \setminus \ln \frac{2}{\ln} \left(x\right) + 1 \setminus \mathrm{dx}$

Next we will split the integral into two:
$\frac{1}{\ln} \left(6\right) \left(\int \setminus \ln \frac{2}{\ln} \left(x\right) \setminus \mathrm{dx} + \int \setminus 1 \setminus \mathrm{dx}\right)$

We know $\int \setminus 1 \setminus \mathrm{dx} = x + C$ and we can take the constant out for the other integral:
$\frac{1}{\ln} \left(6\right) \left(\ln \left(2\right) \int \setminus \frac{1}{\ln} \left(x\right) \setminus \mathrm{dx} + x\right)$

$\int \setminus \frac{1}{\ln} \left(x\right) \setminus \mathrm{dx}$ does not have an elementary solution, but we can represent the answer using the logarithmic integral function, $l i \left(x\right)$:
$\frac{1}{\ln} \left(6\right) \left(\ln \left(2\right) l i \left(x\right) + x\right) + C = \frac{\ln \left(2\right) l i \left(x\right) + x}{\ln} \left(6\right) + C$