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    Question from book

    Given that y=arctan(x-1/x) show that

    dy/dx = (x+1)/(x^4-x^2+1)

    I don't get that - I get

    dy/dx = (x^2+1)/(x^4-x^2+1)

    Can someone do an answer check for me, I've tried it a couple of different methods and my answer differs from the book
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    so annoying, is that arctan((x-1)/x) or arctan(x -1/x)?
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    I agree with your answer.
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    (Original post by Kolya)
    I agree with your answer.
    did you just delete your post then repost the same post?
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    (Original post by Totally Tom)
    did you just delete your post then repost the same post?
    I thought I might have made a mistake so deleted while I checked, but it turned out I was correct all along. :blushing:
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    thanks for that - i need another answer check:

    integrate 1/(x^2)-4

    I've got -1/2artanh(1/2)x

    The book has something with ln in it - I'm guessin they're way off
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    can you make it clearer what the actual integral is?

    is it
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle\int{\frac{1}{x^2-4}
    ?

    if so, I would use partial fractions. then it will have something with ln in it.
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    Yes, it is. I can see why using partial fractions would help, but surely I can also do:

     \frac {1}{x^2-4}

    =
     \frac {1}{-4(1 - (\frac{x}{2})^2})

    and then use a standard result to integrate into

     -\frac{1}{4} 2artanh \frac{x}{2} + c
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    (Original post by studienka)
    Yes, it is. I can see why using partial fractions would help, but surely I can also do:

     \frac {1}{x^2-4}

    =
     \frac {1}{-4(1 - (\frac{x}{2})^2})

    and then use a standard result to integrate into

     -\frac{1}{4} 2artanh \frac{x}{2} + c
    they're equivalent.
 
 
 
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