Why does using IBP with setting say u or dv as 1 work?
- Thread starter Krishang
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blamocur
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Did you look at examples of IBP in your textbooks? Why do you think it shouldn't work?
As for [imath]\int \ln x dx[/imath] : try representing [imath]\ln x dx[/imath] as [imath]u dv[/imath] and see what you might get for [imath]v du[/imath] and [imath]uv[/imath].
As for [imath]\int \ln x dx[/imath] : try representing [imath]\ln x dx[/imath] as [imath]u dv[/imath] and see what you might get for [imath]v du[/imath] and [imath]uv[/imath].
Dr.Peterson
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Please show us an example of what you are asking about, and point out what doesn't work. In particular, show your work for the example you mentioned. Then we'll have a lot more to talk about, rather than a vague, unexplained "doubt" and "feeling".
I have another one, [math]\int \arctan(x)[/math] Here, we set u as [imath]\arctan(x)[/imath] and dv as 1 dx. It just seems like a waste because I am just wanting to integrate this expression and I took u as the expression itself. How does using IBP work here?
Dr.Peterson
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Please show your attempt at carrying out the work; you may either discover how it does work, or reveal some mistake in your understanding of the method.
The point of the method is that you can differentiate the arctan, whereas you can't (yet) integrate it, so you are replacing something you can't do with something you can. That's not unreasonable ...
Yeah that makes sense. Instead of integrating directly, I could differentiate which is usually easier and then I can do integration of dv and then the full integrand net. Thank you
