How do you integrate #lnx/x^.5#?

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Answer 1 · 2016-05-26T02:38:02

#2sqrtx(lnx-2)+C#

This can be written as

#intlnx/sqrtxdx#

#intudv=uv-intvdu#

Here, let #u=lnx# and #dv=1/sqrtxdx#.

These imply that:

#{(u=lnx" "=>" "du=1/xdx),(dv=x^(-1/2)dx" "=>" "v=2x^(1/2)=2sqrtx):}#

Thus,

#intlnx/sqrtxdx=2sqrtxlnx-int(2sqrtx)/xdx#

#=2sqrtxlnx-2intx^(-1/2)dx#

#=2sqrtxlnx-4sqrtx+C#

#=2sqrtx(lnx-2)+C#