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appleduardo Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ dx }{ x ^{2} +2x + 1}\]
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
take \[x ^{2}+2x+1=(x+1)^{2}\] and proceed by substiution
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
did u get it..............
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.0
yeah! but i forgot to type 2 before x^2, :/ so the function is:\[\int\limits_{}^{}\frac{ dx }{ 2x^{2} + 2x +1 }\]
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.0
what i did is this: \[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ dx }{ x ^{2} +2x +1}\] but then i get stuck!
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
then divide and multiply the whole denominator by 2\[2(x ^{2}+x+\frac{ 1 }{ 2 })\] then you need to make it into a square expression as\[x ^{2}+x*2*\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }\]add 1/4 and subtract 1/4 then \[(x+\frac{ 1 }{ 2 })^{2}=x^{2}+x+\frac{ 1 }{ 4 }\]substitute and then solve
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.0
yeah thanks, i will try and ill post when im done :D
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.0
i got: \[2arc tg \frac{ x +0.5 }{ 0.5 } + c\] is that correct?
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.0
(correction)\[arc tg \frac{ x +0.5 }{ 0.5 } + c\]
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
the question takes the form................\[\frac{ 1 }{ (x+\frac{ 1 }{ 2 })^{2}(\frac{ 1 }{ \sqrt{2} })^{2} } \]
 one year ago
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