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(b)=log_x(b)/log_x(a)# where #log_x# can be any logarithm.

We'll simplify to the natural logarithm, since we have that on the bottom:

#int\ log_6(2x)/ln(x)\ dx=int\ (ln(2x)/ln(6))/ln(x)\ dx=int\ ln(2x)/ln(6)*1/ln(x)\ dx#

Now we can take the constant out:

#1/ln(6)int\ ln(2x)/ln(x)\ dx#

We can also use the rule that #log_x(ab)=log_x(a)+log_x(b)# to rewrite in the following way:

#1/ln(6)int\ (ln(2)+ln(x))/ln(x)\ dx=1/ln(6)int\ ln(2)/ln(x)+1\ dx#

Next we will split the integral into two:

#1/ln(6)(int\ ln(2)/ln(x)\ dx+int\ 1\ dx)#

We know #int\ 1\ dx=x+C# and we can take the constant out for the other integral:

#1/ln(6)(ln(2)int\ 1/ln(x)\ dx+x)#

#int\ 1/ln(x)\ dx# does not have an elementary solution, but we can represent the answer using the logarithmic integral function, #li(x)#:

#1/ln(6)(ln(2)li(x)+x)+C=(ln(2)li(x)+x)/ln(6)+C#