Troubleshooting a Simple ln Integration Problem

[igor123d]

I have some stupid trouble with a simple integration. f(x)=ln(x^(1/2))/x

I try using u substitution. u=ln(x^(1/2)) Then du=1/(x^(1/2)*1/(2x^(1/2))dx=dx/(2x)
Then dx should be 2xdu Then plugging back in I should have intg(2u*du) which would give me (ln(x^(1/2))^2; yet the answer my calculator gives me is that answer divided by four; it's like the 2 in the u substitution ends up below xdx, but I don't see how that's possible.

I will greatly appreciate any help.

 

[learningphysics]

I have some stupid trouble with a simple integration. f(x)=ln(x^(1/2))/x

I try using u substitution. u=ln(x^(1/2)) Then du=1/(x^(1/2)*1/(2x^(1/2))dx=dx/(2x)
Then dx should be 2xdu Then plugging back in I should have intg(2u*du) which would give me (ln(x^(1/2))^2; yet the answer my calculator gives me is that answer divided by four; it's like the 2 in the u substitution ends up below xdx, but I don't see how that's possible.

I will greatly appreciate any help.

The calculator is giving you $$\frac{1}{4}(lnx)^2$$ which is the same as your answer. Notice that the calculator gives lnx, not ln(x^(1/2)). It just took the exponent out of the ln.

 

[igor123d]

Stupid Calculator! Lol, thanks a lot. I knew there must have been something really idiotic at the bottom, but being a bit OCD I couldn't rest until I found the cause, and I wasted more than an hour with this problem :). Well thanks again, now I can sleep.