What is $int (lnx) / x^(1/2)dx$ ?

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Answer 1 · 2015-12-15T15:23:01

Put this in your mathematical toolbox: $int x^n lnx dx$ can be done by parts. ( $u=lnx$ and $dv = x^n dx$ )

In this case:

$intx^(-1/2) lnx dx$

Let $u = lnx$ and $dv = x^(-1/2) dx$ .
So that $du = 1/x dx$ and $v = 2x^(1/2)$ .

$uv-vdu = 2x^(1/2)lnx-2 int x^(1/2) 1/x dx$

$= 2x^(1/2)lnx-2 int x^(-1/2)dx$

$= 2x^(1/2)lnx-4x^(1/2) +C$

What is int (lnx) / x^(1/2)dxint (lnx) / x^(1/2)dx?