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anonymous

  • 5 years ago

I know that similar functions can have very different antiderivatives. For example, even though int((x)/((x^4)+1)) dx looks very similar to int(((x^4)+1)/(x)), their solutions are very different. How can I prove that I'm right by solving both?

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  1. amistre64
    • 5 years ago
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    you wanna indefinite integral both of them right?

  2. anonymous
    • 5 years ago
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    Yes.

  3. amistre64
    • 5 years ago
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    the last one just split the fraction: x^4/x +1/x and integrate

  4. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{x^4}{x} + \frac{1}{x} dx\]

  5. anonymous
    • 5 years ago
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    Right...

  6. amistre64
    • 5 years ago
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    the second one we can try subsituting u=x^4+1

  7. amistre64
    • 5 years ago
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    by second I mean first :)

  8. anonymous
    • 5 years ago
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    I gotcha :) I figured as much...first and second...same thing. right? haha!

  9. anonymous
    • 5 years ago
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    so, on the first one you end up with int(1/u)du?

  10. anonymous
    • 5 years ago
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    \[\int\limits 1/u\]

  11. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{1}{4u(u-1)^{1/2}}\]this is what I get :)

  12. amistre64
    • 5 years ago
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    ..du

  13. amistre64
    • 5 years ago
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    put the u under the radial and take out the 1/4

  14. amistre64
    • 5 years ago
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    sounds good, looks bad; decompose the fraction :)

  15. amistre64
    • 5 years ago
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    \[\frac{1}{4}\int\limits_{}\frac{1}{u}du - \int\limits_{} \frac{1}{(u+1)^{1/2}}du\]

  16. amistre64
    • 5 years ago
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    umm... what did I do lol...hold up on that

  17. amistre64
    • 5 years ago
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    yeah, thats right, I couldnt recall where I got (u-1)^1/2 from

  18. anonymous
    • 5 years ago
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    Yeah, I was about to say that you were correct. So, would you mind walking me through how to solve the second one?

  19. amistre64
    • 5 years ago
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    and by second do you mean first?

  20. anonymous
    • 5 years ago
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    Haha! yes, the first one we talked about anyway.

  21. amistre64
    • 5 years ago
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    ok.... whenever we have 2 numbers above a single term; we can split that up to its baser seqments: x^4 + 1 x^4 1 ------- = ---- + ---- right? x x x

  22. amistre64
    • 5 years ago
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    like denominators simply merge and you work the tops; well, they can be split up just as easily

  23. anonymous
    • 5 years ago
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    yes.

  24. amistre64
    • 5 years ago
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    we get: x^3 + 1/x which integrates to: x^4/4 + ln(x) + C

  25. anonymous
    • 5 years ago
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    Oh! I understand! Thank you so much. Now, do we need to add a plus c to the second one we did?

  26. amistre64
    • 5 years ago
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    whenever we have a inegral with no limits attached to it; we have to include the "constant of integration"...so yes. +C is just a place holder till we get more information about it to anchor it to the graph.

  27. amistre64
    • 5 years ago
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    was the decomposition one right? or do we know?

  28. anonymous
    • 5 years ago
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    I don't think we know, unless we can infer that it is from the original problem I gave.

  29. amistre64
    • 5 years ago
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    let me redo that one on paper to see if I got it right....

  30. amistre64
    • 5 years ago
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    running into problems with it ...

  31. amistre64
    • 5 years ago
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    i might just be over thinking it, let me try trig substitution...

  32. anonymous
    • 5 years ago
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    I didn't really notice anything wrong, but I could just be trying to do it wrong. Who knows.

  33. amistre64
    • 5 years ago
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    when I decomped it I allowed the square root to be a negative number... so I had to go another route...

  34. anonymous
    • 5 years ago
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    OH wow.

  35. amistre64
    • 5 years ago
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    if I did it right...gonna have to try to derive it down; \[\frac{\tan^{-1}(x^2)}{2}\]

  36. amistre64
    • 5 years ago
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    I think thats it :)\[Dx (\frac{\tan^{-1}(x^2)}{2}) \rightarrow \frac{2x}{2[(x^2)^2+1]}\]

  37. amistre64
    • 5 years ago
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    which = x/(x^4 +1) :)

  38. amistre64
    • 5 years ago
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    \[\frac{\tan^{-1}(x^2)}{2} + C\]

  39. anonymous
    • 5 years ago
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    Okay...I see I see...Thank you so much for all of your effort and help. :)

  40. amistre64
    • 5 years ago
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    youre welcome :)

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