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In the thumbnail, the denominator should be Bx+C / x²+1 instead of Bx+C / x²+x+1
@@mohannad_139 OH CRAP, I DIDNT EVEN REALIZE THAT 0_0!!! Thanks for letting me know!
Christmas in 8 minutes for me
@@r1ckthe 0o0
Sum of geometric serie formula ?
1=A(1+x^2)+(Bx+C)(1-x)A=B=C=1/2I=1/2•int(1/(1-x)+(x+1)/(1+x^2))dxI=-1/2•ln|1-x|+1/4•ln|1+x^2|+1/2•arctan(x)+C
Did not understand your explanations.
![If you ain't olympiad, DUMMIFY! [MIT Integration Bee 2016]](http://i.ytimg.com/vi/v8S8D1JWxa0/mqdefault.jpg)
![If you ain't olympiad, DUMMIFY! [MIT Integration Bee 2016]](/img/tr.png)
![Putnam Product in an Integral??? [MIT Integration Bee 2015]](http://i.ytimg.com/vi/itLQTzTCedk/mqdefault.jpg)
In the thumbnail, the denominator should be Bx+C / x²+1 instead of Bx+C / x²+x+1
@@mohannad_139 OH CRAP, I DIDNT EVEN REALIZE THAT 0_0!!! Thanks for letting me know!
Christmas in 8 minutes for me
@@r1ckthe 0o0
Sum of geometric serie formula ?
1=A(1+x^2)+(Bx+C)(1-x)
A=B=C=1/2
I=1/2•int(1/(1-x)+(x+1)/(1+x^2))dx
I=-1/2•ln|1-x|+1/4•ln|1+x^2|+1/2•arctan(x)+C
Did not understand your explanations.