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koreanub Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{?}^{?} \frac{ 1 }{ \sqrt{x}(1+x) }\]
 2 years ago

koreanub Group TitleBest ResponseYou've already chosen the best response.0
The question marks are not suppose to be there, so it's the problem without the question marks.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\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.
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
i believe usub will work
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
\[u=\sqrt{x}\] ha?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
woops!! that also works!!
 2 years ago

koreanub Group TitleBest ResponseYou've already chosen the best response.0
For this exercise, I'm technically not suposed to use partial fractions
 2 years ago

koreanub Group TitleBest ResponseYou've already chosen the best response.0
I may be missing something, but I can't seem to find the answer with u = \[\sqrt{x} \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
change all x's into u's probably you would end up with 1/1+u^2 or something like that.
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
You have to mess with the U a little bit kore :) but it will work.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
x= tan^2 theta might work
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
this is same as using trig subs x = tan^2 theta in original.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
but lol ... we know int 1/1+x^2 = arctan (x) which is one of the standard integral.
 2 years ago

koreanub Group TitleBest ResponseYou've already chosen the best response.0
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. .
 2 years ago
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