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appleduardo

  • 2 years ago

whats the integral of the following function?

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  1. appleduardo
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ dx }{ x ^{2} +2x + 1}\]

  2. harsh314
    • 2 years ago
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    take \[x ^{2}+2x+1=(x+1)^{2}\] and proceed by substiution

  3. harsh314
    • 2 years ago
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    did u get it..............

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

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

  6. harsh314
    • 2 years ago
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    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

  7. appleduardo
    • 2 years ago
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    yeah thanks, i will try and ill post when im done :D

  8. appleduardo
    • 2 years ago
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    i got: \[2arc tg \frac{ x +0.5 }{ 0.5 } + c\] is that correct?

  9. appleduardo
    • 2 years ago
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    (correction)\[arc tg \frac{ x +0.5 }{ 0.5 } + c\]

  10. harsh314
    • 2 years ago
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    the question takes the form................\[\frac{ 1 }{ (x+\frac{ 1 }{ 2 })^{2}-(\frac{ 1 }{ \sqrt{2} })^{2} } \]

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