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LACHEEK989 Group Title

∫1/x2+x+1

  • one year ago
  • one year ago

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  1. wio Group Title
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    Find the roots, then you can factor it. Then you want to try partial fraction decomposition.

    • one year ago
  2. LACHEEK989 Group Title
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    I am lost at completing the square.

    • one year ago
  3. zepdrix Group Title
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    \[\large x^2+bx\]To complete the square, we take half of the \(\large b\) term, and square it. \[\large x^2+bx+\left(\frac{b}{2}\right)^2\]Now we can't just add this number, it will change the value of our polynomial, we want to keep it balanced. So we have to also subtract this number. \[\large \color{royalblue}{x^2+bx+\left(\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]The reason for doing this is, the blue part will now factor down into a perfect square. A binomial containing x and half of the b term.\[\large \color{royalblue}{\left(x+\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]

    • one year ago
  4. zepdrix Group Title
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    Let's see if we can apply this to our problem here.

    • one year ago
  5. zepdrix Group Title
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    \[\large x^2+x+1\]The \(\large b\) coefficient appears to be 1. (The middle term ~ because \(\large x\) is the same as \(\large 1\cdot x\) ). So to to complete the square, we'll take half of 1, and square it.\[\large \color{orangered}{x^2+x+\left(\frac{1}{2}\right)^2}-\left(\frac{1}{2}\right)^2+1\]I moved the 1 out of the way, we don't want to deal with that right now. Ok the orange part should give us a perfect square now,\[\large \color{orangered}{\left(x+\frac{1}{2}\right)^2}-\left(\frac{1}{2}\right)^2+1\] Which will simplify further to,\[\large \left(x+\frac{1}{2}\right)^2+\frac{3}{4}\]

    • one year ago
  6. LACHEEK989 Group Title
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    \[4/3 \int\limits_{}^{}1/((4/3) u^2 +1) du\] What next?

    • one year ago
  7. LACHEEK989 Group Title
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    (3/4)∫1/((4/3)u2+1)du I think this is right. What do I do next?

    • one year ago
  8. Outkast3r09 Group Title
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    this seems much more easier than what is being done... but thats just me

    • one year ago
  9. Outkast3r09 Group Title
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    \[\frac{4}{3}\int\frac{1}{\frac{4}{3}u^2+1}du=\frac{\frac{4}{3}}{\frac{4}{3}} \int \frac{1}{u^2+1}du\]

    • one year ago
  10. LACHEEK989 Group Title
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    I agree. the book (answer in the back) and wolfram alpha I believe use a substitution here of 2/sqrt3 which gets complicated..

    • one year ago
  11. Outkast3r09 Group Title
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    oh wait

    • one year ago
  12. Outkast3r09 Group Title
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    is it \[\frac{1}{4/3(u^2+1)}\]

    • one year ago
  13. Outkast3r09 Group Title
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    ?

    • one year ago
  14. LACHEEK989 Group Title
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    1/x2+x+1 tp begin with.

    • one year ago
  15. LACHEEK989 Group Title
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    Does that answer your question?

    • one year ago
  16. Outkast3r09 Group Title
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    what is problem... like that is in the book

    • one year ago
  17. Outkast3r09 Group Title
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    without anything done to it

    • one year ago
  18. LACHEEK989 Group Title
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    \[\int\limits_{}^{} (x^2 -x + 2) / (x^3 -1) \ \ dx\]

    • one year ago
  19. LACHEEK989 Group Title
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    I understand the partial fraction decomposition element to this. I am still lost at the point mentioned to complete the square. I haven't been taught yet.

    • one year ago
  20. Outkast3r09 Group Title
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    alright well whered the partials you used?

    • one year ago
  21. Outkast3r09 Group Title
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    I ask this because if you use partials you should've gotten a natural log for one of the fractions

    • one year ago
  22. wio Group Title
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    When you have \(u^2+1\) it's a sign to try the trig sub \(u=\tan\theta\).

    • one year ago
  23. LACHEEK989 Group Title
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    yeah the answer so far is .. \[2/3 \ln (x+1) + 1/6 \ln (x^2 + x + 1) -9/2\int\limits_{}^{} dx / (x^2 + x + 1) \]

    • one year ago
  24. LACHEEK989 Group Title
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    u = 4/sqrt3 ?

    • one year ago
  25. LACHEEK989 Group Title
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    sorry 2/sqrt3 = u for the substitution?

    • one year ago
  26. wio Group Title
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    What is the original integral?

    • one year ago
  27. LACHEEK989 Group Title
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    ∫(x2−x+2)/(x3−1) dx

    • one year ago
  28. Outkast3r09 Group Title
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    the last part you have to

    • one year ago
  29. Outkast3r09 Group Title
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    complete the square and use trig sub

    • one year ago
  30. Outkast3r09 Group Title
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    \[x^2+x+1\] half of b = 1/2.. \[\frac{b}{4}^2=(\frac{b}{2})^2=\frac{1}{4}\]

    • one year ago
  31. Outkast3r09 Group Title
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    if you add that you must subtract \[\frac{1}{x^2+x+\frac{1}{4}+1-\frac{1}{4}}\]

    • one year ago
  32. Outkast3r09 Group Title
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    this is where he ot \[\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}}\]

    • one year ago
  33. LACHEEK989 Group Title
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    then factor out the 3/4 in the bottom.

    • one year ago
  34. Outkast3r09 Group Title
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    ok now you get to where you were before right?

    • one year ago
  35. LACHEEK989 Group Title
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    right

    • one year ago
  36. JenniferSmart1 Group Title
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    :'(

    • one year ago
  37. Outkast3r09 Group Title
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    now let \[x=\frac{a}{b}tan(\theta)\]

    • one year ago
  38. LACHEEK989 Group Title
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    is a/b 3/4 or does should I do a substitution like 2/sqrt 3 ?

    • one year ago
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