LN integration problem

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ninguen

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Jan 1, 2012
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I'm working on a problem where Mathway and Shaum's give 2 very different answers... and the Mathway explanation is hard to understand.

∫ √(ln(x+3))/(x) dx

So far, I've figured out that:

∫ (1/2)(ln(x+3))/(x) dx

U = ln(x+3)
du= 1/(x+3) dx

du(x+3) = dx

What now?
 
Is this your integral. This is what you have written in the first line:

\(\displaystyle \displaystyle \int\frac{\sqrt{ln(x+3)}}{x}dx\)?.

If so, I do not think this has a closed form. What is Schaum and Mathway giving?.

Or is it \(\displaystyle \displaystyle \int \frac{ln(\sqrt{x+3})}{x}dx=\int\frac{ln(x+3)}{2x}dx\)

Are there limits of integration given, by chance?.

The latter would still be challenging to integrate with no limits of integration.

As is, it could be expressed in terms of what is known as the dilogarithm. Rather beyond someone self-teaching themselves calculus.

Now, if it were \(\displaystyle \displaystyle \int ln\left(\frac{\sqrt{x+3}}{x}\right)dx=\frac{1}{2}\int ln(x+3)dx-\int ln(x)dx\), then we would be in business.

This is integrable.
 
Last edited:
My original formula is right out of the book.

Schaum's says the answer should be (2/3) ln(x+3)^(3/2)

It even gives a hint: "Use quick formula I"
 
ninguen said:
My original formula is right out of the book.
Schaum's says the answer should be
(2/3) [ln(x+3)]^(3/2)
Click to expand...
That is the answer to \(\displaystyle \displaystyle\int {\frac{{\sqrt {\ln (x + 3)} }}{{(x + 3)}}dx} \)
 
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