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mony01

  • 2 years ago

Anyone know how to do this integral?

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  1. mony01
    • 2 years ago
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    \[\int\limits \frac{ x ^{2} +x+1}{ (x ^{2}+1)^{2} }dx\]

  2. wio
    • 2 years ago
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    Play around with the fraction.

  3. wio
    • 2 years ago
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    Maybe partial fraction decomposition

  4. myininaya
    • 2 years ago
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    did you try a trig sub i think that would work just fine

  5. mony01
    • 2 years ago
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    you think i can try sin?

  6. myininaya
    • 2 years ago
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    I think you can try tan

  7. myininaya
    • 2 years ago
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    the hint was the x^2+1 on bottom

  8. myininaya
    • 2 years ago
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    tan^2(theta)+1=sec^2(theta)

  9. mony01
    • 2 years ago
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    would the set up be integral x^2+x+1/sex^2 theta

  10. myininaya
    • 2 years ago
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    well you have to replace all the x's and the dx

  11. myininaya
    • 2 years ago
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    and also you are leaving off the square on bottom

  12. myininaya
    • 2 years ago
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    \[x=\tan(\theta) => dx=\sec^2(\theta) d \theta \] Replace all the x's with tan(theta) Replace the dx with sec^2(theta) d theta

  13. mony01
    • 2 years ago
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    is it integral tan^2theta+tan theta+1/sec^2 theta d (theta)

  14. myininaya
    • 2 years ago
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    If I think what you wrote is what I think then yes. You mean the following I assume: integral of (tan^2(theta)+tan(theta)+1)/sec^2(theta) d(theta)

  15. myininaya
    • 2 years ago
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    \[\int\limits_{}^{}\frac{\tan^2(\theta)}{\sec^2(\theta)} d \theta +\int\limits_{}^{}\frac{\tan(\theta)}{\sec^2(\theta)} d \theta +\int\limits_{}^{} \frac{1}{\sec^2(\theta) }d \theta \]

  16. myininaya
    • 2 years ago
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    Look at them three separately

  17. gorv
    • 2 years ago
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    \[\int\limits_{}^{}(\frac{ x^2+1 }{ (x^2+1)^2 } +\frac{ x }{ (x^2+1)^2 })*dx\]

  18. gorv
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ dx }{ x^2+1 } +\int\limits_{}^{}\frac{ x }{ (x^2+1)^2 }\]

  19. gorv
    • 2 years ago
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    first one is standard formula

  20. mony01
    • 2 years ago
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    how can i figure out the answer?

  21. gorv
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ x }{ (x^2+1)^2 } *dx\] for this x^2+1=t 2xdx=dt xdx=dt/2

  22. gorv
    • 2 years ago
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    \[\int\limits_{}^{} \frac{ dt }{ 2*t^2 }\]

  23. gorv
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ dx }{ x^2+1 } +\int\limits_{}^{}\frac{ dt }{ 2t^2 }\]

  24. gorv
    • 2 years ago
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    can u solve it @mony01

  25. mony01
    • 2 years ago
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    is it sec^2 (theta)/2(x^2+1)^2

  26. myininaya
    • 2 years ago
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    I do like gorv's way. But either way is fine. gorv's is simpler though.

  27. gorv
    • 2 years ago
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    actual its less calculative...both ways anns will be same

  28. mony01
    • 2 years ago
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    how can i solve the rest of it?

  29. myininaya
    • 2 years ago
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    What question on what part do you have?

  30. mony01
    • 2 years ago
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    \[\int\limits \frac{ dx }{ x ^{2}+1}+\int\limits \frac{ dt }{ 2t ^{2} }\]

  31. myininaya
    • 2 years ago
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    the first integral is just remembering that it is acran(x) the second one 1/2 is a constant multiple and you should know how to integrate t^(-2) at this point in calculus

  32. mony01
    • 2 years ago
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    is the answer arctan(x)-1/2(x^2+1)+C

  33. myininaya
    • 2 years ago
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    Yes that's right you mean arctan(x)- 1/[2(x^2+1)] +C good job

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