integrate 1/√-logx within limits 0 to 1 with respect to x

Dear Student

Substitute=−ln(x)
⟶du/dx=−1/x
⟶dx=−xdu,

Now solving:
∫(e^−u/√u)du

∫(2e^{−v^2} / √π)dv

This is special function called Gauss error function.
In this question you’ll get to use integration of gauss error function which is not in school level.

Thanks