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Here's slightly different approach
We have
I=∫10−lnx−−−−−√dx=∫10[ln1x]1/2dx
By using IBP
[ln1x]1/2=u⟺du=−12x[ln1x]−1/2dx
dx=dv⟺x=v
I=x[ln1x]1/2∣∣∣10+12∫10[ln1x]−1/2dx=12∫10[ln1x]−1/2dx
Now by substututing
t2=ln1x⟹e−t2=x⟺dx=−2te−t2dt
We get
I=∫∞0e−t2dt=π−−√2
Hence,
∫10−lnx−−−−−√dx=π−−√2
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