Show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C

  • MHB
  • Thread starter bergausstein
  • Start date
  • Tags
    Dx
Now, we can use integration by parts:\int\log_b x\,dx=x\log_b x-\int x\left(\frac{\ln x}{\ln b}\right)'dx\\=x\log_b x-\int \frac{1}{\ln b}\,dx\\=x\log_b x-\frac{x}{\ln b}+C.In summary, using the change of base theorem and integration by parts, we can show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C.
  • #1
bergausstein
191
0
Show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C

can you show me the complete solution to this prob? thanks!
 
Physics news on Phys.org
  • #2
We typically do not provide full solutions, but offer help so that those posting questions are able to work the problem for themselves, thereby gaining more. I know I learn more by doing than by watching. (Nerd)

I would recommend using the change of base theorem on the integrand first:

\(\displaystyle \log_b(x)=\frac{\ln(x)}{\ln(b)}\)

Then, I would use integration by parts, after you pull the constant from the integrand. Can you proceed?

If you get stuck or need clarification, please feel free to show what you have tried and where you are stuck, and we will be glad to help.
 
  • #3
bergausstein said:
Show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C

can you show me the complete solution to this prob? thanks!
From the laws of logarithms:\(\displaystyle \log_b x=\frac{\log_c x}{\log_c b}\)

So if we select \(\displaystyle c=e\,\), and use Naperian/Natural logarithms, then this becomes

\(\displaystyle \log_b x=\frac{\ln x}{\ln b}\)
So your integral would be\(\displaystyle \int \log_b x\,dx=\frac{1}{\ln b}\int \ln x\, dx=\)\(\displaystyle \frac{1}{\ln b}\left[x\ln x-\int x\frac{1}{x}\,dx\right]=\)\(\displaystyle \frac{1}{\ln b}\left[x\ln x-\int \,dx\right]=\)\(\displaystyle \frac{1}{\ln b}\left[x\ln x-x\right]+C=\)\(\displaystyle x\left(\frac{\ln x}{\ln b}-\frac{1}{\ln b}\right)+C=\)\(\displaystyle x\left(\log_b x-\frac{1}{\ln b}\right)+C\)
 

FAQ: Show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C

1. What does the notation "∫ logb(x) dx" mean?

The notation "∫ logb(x) dx" represents the indefinite integral of the logarithmic function with base b. It is the inverse operation of finding the derivative of a function.

2. What is the value of C in the equation "∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C"?

C is the constant of integration and its exact value depends on the initial conditions of the problem. It is added to the result of the indefinite integral to account for all possible solutions.

3. How do you show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C?

One way to show this is by using the power rule for integration and the logarithmic properties. By rewriting logb(x) as ln(x) / ln(b), we can apply the power rule and solve for the integral. The constant of integration can also be added to the final result.

4. What is the significance of the ln(b) term in the equation "∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C"?

The ln(b) term is the natural logarithm of the base b. It appears in the result of the integration because the logarithmic function is defined using the natural logarithm. It also plays a crucial role in the properties of logarithms and their derivatives.

5. Can this equation be applied to any base b or only certain values?

This equation can be applied to any base b as long as the logarithmic function is defined for that base. However, the properties of logarithms and their derivatives may vary for different bases and may require different methods of integration.

Similar threads

Replies
15
Views
2K
Replies
31
Views
2K
Replies
5
Views
1K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
Back
Top