How Do You Integrate (x^(1/2))/ln(x) Using Integration by Parts?

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QUESTION:

The question is to find the improper integral of (x^1/2)/lnx dx.

MY ATTEMPT:
1)I tried it byparts, by taking 1/ln x as 'u' or the first function but i got stuck.

2)Alternatively, I tried substituting x=e^2t in hopes to eliminate ln for a simpler byparts integration, but that didn't work
out.

Please help asap. Thanks.
 
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Engineerbrah said:
QUESTION:

The question is to find the improper integral of (x^1/2)/lnx dx.

MY ATTEMPT:
1)I tried it byparts, by taking 1/ln x as 'u' or the first function but i got stuck.

2)Alternatively, I tried substituting x=e^2t in hopes to eliminate ln for a simpler byparts integration, but that didn't work
out.

Please help asap. Thanks.
You can apply parts directly. Apply parts in the form such that the evaluation of the new integral involves the derivative of (1/lnx).
 
Devin said:
You can apply parts directly. Apply parts in the form such that the evaluation of the new integral involves the derivative of (1/lnx).

I tried it this way. The furthest I got was

(2*(x)^(1/2))/lnx - Integral of (2/((x)^(1/2))(lnx)^2 dx

Still not able to attain the answer.
 
Engineerbrah said:
I tried it this way. The furthest I got was

(2*(x)^(1/2))/lnx - Integral of (2/((x)^(1/2))(lnx)^2 dx

Still not able to attain the answer.
Devin said:
You can apply parts directly. Apply parts in the form such that the evaluation of the new integral involves the derivative of (1/lnx).
I believe this to be the way to approach. Then the evaluation of the new integral is straightforward.

<Mod note: attachment deleted>
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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