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Irrational natural log integral

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data
    the indefinite integral of (1+lnx)^(1/2)/(xlnx) dx


    2. Relevant equations
    n/a


    3. The attempt at a solution
    There aren't any x^2 in the root sign, so I don't think it can be a trig substitution. The only logical u sub I see is to let u=lnx. In that case, du=dx/x so the integral becomes (u+1)^(1/2)/u du. Unfortunately, I don't have any good ways to evaluate that integral.

    I wanted to try another sub letting m=(u+1)^(1/2) which makes m^2=u+1. Then u=m^2-1 and du=2mdm. that left me with 2 times the integral of m^2/(m^2-1) dm. I think I can solve that using partial fractions [integral of 1+1(m^2-1) dm : A(m-1)+B(m+1)=1, A=-1/2 and B=1/2, solution=m-1/2lnm+1/2lnm, backsub a few times to get (lnx+1)^(1/2)-(1/2)ln(lnx+1)+(1/2)ln(lnx+1)+c] but I'm not at all confident in that answer or in my methods there at the last.
     
  2. jcsd
  3. May 13, 2007 #2
    My first thought was the substitution u = xln(x), then du = (1 + ln(x)) dx, but the square root there messes it up.
     
  4. May 14, 2007 #3

    Dick

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    Homework Helper

    Nice try. You've basically got it. Except that after you finished the m integration you substituted ln(x) for m in the log parts of the expression. m=sqrt(ln(x)+1). And don't forget your overall factor of 2.
     
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