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

bergausstein

Active member
Jul 30, 2013
191
Show that ∫ logb(x) dx = x ∙ ( logb(x) - 1 / ln(b) ) + C

can you show me the complete solution to this prob? thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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


\(\displaystyle \int \log_b x\,dx=\frac{1}{\ln b}\int \ln x\, dx=\)


\(\displaystyle \frac{1}{\ln b}\left[x\ln x-\int x\frac{1}{x}\,dx\right]=\)


\(\displaystyle \frac{1}{\ln b}\left[x\ln x-\int \,dx\right]=\)


\(\displaystyle \frac{1}{\ln b}\left[x\ln x-x\right]+C=\)


\(\displaystyle x\left(\frac{\ln x}{\ln b}-\frac{1}{\ln b}\right)+C=\)


\(\displaystyle x\left(\log_b x-\frac{1}{\ln b}\right)+C\)