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sjg13e

  • one year ago

Integration by Parts problem

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  1. sjg13e
    • one year ago
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    \[\int\limits_{}^{}\tan^{-1}xdx\]

  2. anonymous
    • one year ago
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    try u = arctan x dv = dx

  3. sjg13e
    • one year ago
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    \[\let u = \tan^{-1} and du = \frac{ -1 }{ x^2 +1 }dx\]

  4. sjg13e
    • one year ago
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    then dv = dx and v = x

  5. sjg13e
    • one year ago
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    so, \[\tan^{-1} \frac{ 1 }{ x } x - \int\limits_{}^{}xdx = \tan^{-1} \frac{ 1 }{ x } x - .5x^2 \]

  6. anonymous
    • one year ago
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    * du should be positve du 1/(x²+1)

  7. sjg13e
    • one year ago
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    why is du positive?

  8. anonymous
    • one year ago
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    that's the derivative of arctan x. \[\frac{ d }{ dx }\tan^{-1} x=\frac{ 1 }{ 1+x^2 }\]

  9. sjg13e
    • one year ago
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    yes but chain rule

  10. sjg13e
    • one year ago
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    we need to take the derivative of the inside function, which is 1/x

  11. anonymous
    • one year ago
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    no. that's an inverse function, not an exponent

  12. sjg13e
    • one year ago
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    \[(1/x)\prime = -1/x^2\]

  13. sjg13e
    • one year ago
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    i think it's treated as an exponent

  14. sjg13e
    • one year ago
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    1/x = x^-1 = -1/2 * x^2

  15. sjg13e
    • one year ago
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    ohh I'm sorry!! i forgot to properly write the initial problem. It's supposed to say \[\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x }) dx\]

  16. sjg13e
    • one year ago
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    i apologize. that's totally my fault

  17. anonymous
    • one year ago
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    oh ok

  18. sjg13e
    • one year ago
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    okay, ahah, so sorry about that! so besides that little mishap, does everything else look right?

  19. anonymous
    • one year ago
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    yeah it looks good

  20. sjg13e
    • one year ago
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    okay thank you!

  21. anonymous
    • one year ago
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    yw

  22. sjg13e
    • one year ago
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    Just for future reference. My previous answer was wrong, here's the correct, complete solution: \[\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x })dx = x \tan^{-1}(\frac{ 1 }{ x }) - \int\limits_{}^{}x*\frac{ 1 }{ x^2 +1 }dx \] at this point, you have to do u-subsition in the integral. so, let u = x^2 + 1, which will make .5du = xdx we can bring the - and the .5 out front to give: \[x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ du }{ u }\] \[ =x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\ln|u| \] \[= x \tan^{-1} (\frac{ 1 }{ x }) + \ln |x^2 + 1| + C\]

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