What is the indefinite integral of ${ln(x)}^2$ ?

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Answer 1 · 2016-07-03T20:08:07

$= x ln^2 x -2 x ln x +2 x + C$

$int dx qquad ln^2(x)$

we use IBP - $int u v' = uv - int u'v$

$u = ln^2(x), u' = 2 ln x *1/x$
$v' = 1, v = x$

$implies x ln^2 x - int dx qquad 2 ln x$

$implies x ln^2 x -2 color{red}{ int dx qquad ln x} qquad circ$

for the red bit again we IBP

here

$u = ln x, u' = 1/x$
$v' = 1, v = x$

$implies int dx qquad ln x = x ln x - int dx qquad x*1/x$
$= x ln x - int dx$

$= x ln x - x + C qquad square$

putting $square$ into $circ$

$implies x ln^2 x -2 (x ln x - x ) + C$

$= x ln^2 x -2 x ln x +2 x + C$

What is the indefinite integral of {ln(x)}^2{ln(x)}^2?