Irrational natural log integral

mwaso
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Homework Statement


the indefinite integral of (1+lnx)^(1/2)/(xlnx) dx


Homework Equations


n/a


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.
 
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My first thought was the substitution u = xln(x), then du = (1 + ln(x)) dx, but the square root there messes it up.
 
mwaso said:

Homework Statement


the indefinite integral of (1+lnx)^(1/2)/(xlnx) dx

Homework Equations


n/a

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.

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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