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#### bergausstein

##### Active member

- Jul 30, 2013

- 191

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

- Thread starter bergausstein
- Start date

- Thread starter
- #1

- Jul 30, 2013

- 191

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

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- #2

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.

- Sep 16, 2013

- 337

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

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

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

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

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

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

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

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