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Antiderivative Help

  1. Jan 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the following integral:
    b1719ffb92379d42d595382b30430c6f.png

    2. Relevant equations
    N/A

    3. The attempt at a solution
    b1719ffb92379d42d595382b30430c6f.png
    728c33c9e6b78570668463ea8929d8f2.png
    36a8f99e810595ddc5eef07ea876eefe.png
    from this point I tried a u-substitution by letting u = -3 + 4/x -1/x^2 but this seemed to fail.

    Are any suggestions possible?
     
  2. jcsd
  3. Jan 24, 2016 #2

    SteamKing

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    Try factoring the quadratic under the root sign first. You should be able to find two linear factors.
     
  4. Jan 24, 2016 #3
    Alright thank you will attempt that now.
     
  5. Jan 24, 2016 #4
    Alright I have the indefinite integral factored for the argument however I am unsure how to proceed.
    360ea395d27c22bc3e3818285928f822.png
    I'm going to let u = x-1 which means x = u + 1 and for the second factor (3x-1) = (3x - 1 - 2 + 2) = [3(x-1) + 2]
     
  6. Jan 24, 2016 #5
    I may not be approaching this the right direction since after attempting this method I seem to be unable to proceed
     
  7. Jan 24, 2016 #6
    There's usually several substitutions that work for these types of problems. The simplest one seems to be ##t=1/x##. At least that seem to work for me (but I did this rather quickly so I may have done something wrong).

    Another approach that should work would be is to complete the square and then use the standard substitution ##\arcsin## but in this case it seems to be rather complicated and you would need another substitution ##s=\tan t/2## after as well.
     
  8. Jan 31, 2016 #7
    That looks like I would complete the square of the term under the square root to get something to match arcsec or arccsc. 1/(u √ (u2-a2) is the one I believe i am thinking of..
     
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