Here's the question you clicked on:
koreanub
Integration
\[\int\limits_{?}^{?} \frac{ 1 }{ \sqrt{x}(1+x) }\]
The question marks are not suppose to be there, so it's the problem without the question marks.
\int just use like this for integration ... seems like trig subs seems to work also changing 1 + x = (1 + i sqrt(x))(1 - i sqrt(x)) and then taking partial fraction might work.
i believe u-sub will work
woops!! that also works!!
For this exercise, I'm technically not suposed to use partial fractions
I may be missing something, but I can't seem to find the answer with u = \[\sqrt{x} \]
change all x's into u's probably you would end up with 1/1+u^2 or something like that.
You have to mess with the U a little bit kore :) but it will work.
x= tan^2 theta might work
this is same as using trig subs x = tan^2 theta in original.
but lol ... we know int 1/1+x^2 = arctan (x) which is one of the standard integral.
I got it! Yes!!! I was substituting the first \[\sqrt{x} \] in the integrand. Then, I realized that it gets canceled out. Thank you very much everyone! ... If it wasn't for you, I probably would have spent another 10 minutes on this. -.-