Question

whats the integral of the following function?

Answer 1

\[\int\limits_{}^{}\frac{ dx }{ x ^{2} +2x + 1}\]

Answer 2

take \[x ^{2}+2x+1=(x+1)^{2}\] and proceed by substiution

Answer 3

did u get it..............

Answer 4

yeah! but i forgot to type 2 before x^2, :/ so the function is:\[\int\limits_{}^{}\frac{ dx }{ 2x^{2} + 2x +1 }\]

Answer 5

what i did is this: \[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ dx }{ x ^{2} +2x +1}\] but then i get stuck!

Answer 6

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

Answer 7

yeah thanks, i will try and ill post when im done :D

Answer 8

i got: \[2arc tg \frac{ x +0.5 }{ 0.5 } + c\] is that correct?

Answer 9

(correction)\[arc tg \frac{ x +0.5 }{ 0.5 } + c\]

Answer 10

the question takes the form................\[\frac{ 1 }{ (x+\frac{ 1 }{ 2 })^{2}-(\frac{ 1 }{ \sqrt{2} })^{2} } \]