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ggrree Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{} {dx \over 1+ \sqrt x}\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\int\limits_{}^{}\frac{1}{1+\sqrt{x}} dx\] \[\text{ let } \sqrt{x}=\tan^2(\theta) \] \[ x=\tan^4(\theta)\] => \[dx=4 \tan^3(\theta) \sec^2(\theta) d \theta\] So we have \[\int\limits_{}^{}\frac{4 \tan^3(\theta) \sec^2(\theta)}{1+\tan^2(\theta)} d \theta\] \[4 \int\limits_{}^{}\tan^3(\theta) d \theta\] That should help...
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
We could try another way if you don't like this one...
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits\frac{1}{1+\sqrt{x}} dx\] \[x = u^2 =>u=\sqrt{x} \] \[\frac{dx}{du} =2u =>dx=2u*du\] \[=>\int\limits\frac{1}{1+u} *2u*du =\int\limits\frac{u}{u+1} \] Might be easier..
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
yep that's the one
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
woops, sorry
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
woops, i made a typo, the last one should be.. \[2\int\limits\frac{u}{u+1} du\]
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Let u = sqrtx > u^2 = x => 2udu = dx > 2 Int ( udu/ u + 1) > 2 Int [ 1  1/(u + 1 )] du = 2 [ u  ln(u + 1) ] = 2 ( sqrtx  ln(sqrtx + 1 ) + C
 2 years ago

ggrree Group TitleBest ResponseYou've already chosen the best response.0
thanks guys!
 2 years ago
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