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,

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

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

$=2sqrtxlnx-4sqrtx+C$

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

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