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Integral of 1/[(x^2)*(sqrt(6x+1))]

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int\frac{dx}{x^{2}*\sqrt{6x+1}}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    t=sqrt(6x+1)
    so x=t[itex]^{2}[/itex]-1
    dx=2t dt

    so new integral is

    2*[itex]\int\frac{t}{(t^{2}-1)^{2}*t}[/itex]

    the t's cancel and then i can turn the denominator into (t+1)[itex]^{2}[/itex]*(t-1)[itex]^{2}[/itex]

    By partial fractions I get the new integral to be

    (1/4)*( [itex]\int\frac{dt}{t+1}[/itex]+[itex]\int\frac{dt}{(t+1)^{2}}[/itex]-[itex]\int\frac{dt}{(t-1)}[/itex]+[itex]\int\frac{dt}{(t-1)^{2}}[/itex] )

    so the first and third are easy. The second I make a substitution equal to t+1. The 4th I make a substitution equal to t-1.

    After finding all the integrals I replace all the 't's with (sqrt(6x+1)) and my final answer is


    (1/4)*((ln|[itex]\sqrt{6x+1}[/itex]+1|)-(ln|[itex]\sqrt{6x+1}[/itex]-1|)-[itex]\frac{1}{\sqrt{6x+1}+1}[/itex]-[itex]\frac{1}{\sqrt{6x+1}-1}[/itex])

    But apparently it's wrong and after checking over it I can't find out where. I am getting super frustrated because even when I KNOW the procedure for a problem, I am more often wrong than right. Does anyone also have any general tips to prevent this?
     
  2. jcsd
  3. Feb 7, 2012 #2

    SammyS

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    I haven't checked the whole thing yet, but if
    [itex]\displaystyle t=\sqrt{6x+1}[/itex]​
    then
    [itex]\displaystyle x=\frac{t^2-1}{6}[/itex]​
     
  4. Feb 7, 2012 #3
    OK thanks. That's incredibly frustrating that I made such a dumb mistake but at least I know now
     
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