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can any one help me!!!!!

Mathematics
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|dw:1327218851048:dw|
its calculus
is it t+1 in the bracket

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Other answers:

no its inside log
Is that \[ \int_1^x \frac{B\cdot \log(1+t)}{1+t^2} dt \]
\[\int\limits_{0}^{1} \frac{8\ln(1+t)}{1+t^2} \]
its this ^
This looks like a pretty nasty integral, what class is this for? :)
My first thought would be integration by parts, actually.
ok now i got it
@jem: i am a high school student :P
@was : can u help!!!!
help
ok i wud go in this way by integration by parts take 8 out now u left with (\[8\int\limits_{x}^{1}(\ln(1+t)/(1+t^2)\] now take 1/(1+t^2) as part to integrate , and derivative ln (1+t)
\[\pi \log(2)\]
~~2.17759
integral of 1/(1+t^2) is tanx
arctan^
yehh jemurray typo
@aron: hw did you got the ans ? can you show the steps ? @was:then?
dhash i have simplified enuf lol
i got '0=0' thats why i am asking you
use integration by parts, i have told u what to integrate and what to differentiate ,
i didnot get the ans in that way
You've neglected to mention the integral \[ \int_0^1 \frac{\tan^{-1}(x)}{1+x} dx \] which is pretty damn complicated, I would say.
4 i (-Li_2((1/2-i/2) (t+1))+Li_2((1/2+i/2) (t+1))+log(t+1) (log((-1/2-i/2) (t+i))-log(1-(1/2-i/2) (t+1))))+constant
yep jem is right
Now, just plug in the limits.
With respect Aron, we all know how to use wolfram alpha.
haha jem
jem bt what else cud b easier way other than by parts to teach a high scool student
you're missing my point, integration by parts still requires you to solve that integral I wrote, which is kinda rough.
i know jemm im tryin to simplify it
Try x = tan y
yeah got it place t =arctany then its simple
cool thanks that works

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