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

Answer 1

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

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#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To integrate ( f(x) = \frac{\log_6(2x)}{\ln(x)} ) using the quotient rule, we first express it as the difference of two functions, ( u(x) ) and ( v(x) ), where ( u(x) = \log_6(2x) ) and ( v(x) = \ln(x) ). Then, we apply the quotient rule, which states that the derivative of ( \frac{u(x)}{v(x)} ) is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ).

First, we find the derivatives of ( u(x) ) and ( v(x) ). The derivative of ( \log_6(2x) ) with respect to ( x ) is ( \frac{1}{\ln(6)} \cdot \frac{1}{2x} ) by the chain rule. The derivative of ( \ln(x) ) is ( \frac{1}{x} ).

Now, we apply the quotient rule:

[ f'(x) = \frac{\left(\frac{1}{\ln(6)} \cdot \frac{1}{2x}\right) \cdot \ln(x) - \log_6(2x) \cdot \frac{1}{x}}{(\ln(x))^2} ]

[ = \frac{\frac{1}{2x \cdot \ln(6)} \cdot \ln(x) - \frac{\log_6(2x)}{x}}{(\ln(x))^2} ]

[ = \frac{\ln(x) - \log_6(2x) \cdot \ln(x)}{2x \cdot \ln(6) \cdot (\ln(x))^2} - \frac{\log_6(2x)}{x \cdot (\ln(x))^2} ]

This is the derivative of ( f(x) ). To find the integral of ( f(x) ), we integrate ( f'(x) ) with respect to ( x ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7