Why does using IBP with setting say u or dv as 1 work?

[Krishang]

I have had this doubt for quite some time as on the surface it just feels like it shouldn't work. It's like as if I am just rewriting the integral. How does this work in cases like say for [imath] \int \ln(x) [/imath]?

 

[blamocur]

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].