How do you find the integral of $x {(ln x)}^{3} d x$ $x(ln x)^3 dx$ ?

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Answer 1 · 2015-06-30T21:31:13

Let:
$u = {ln}^{3} x$ $u = ln^3x$
$d u = \frac{3 {ln}^{2} x}{x} d x$ $du = (3ln^2x)/xdx$
$d v = x d x$ $dv = xdx$
$v = \frac{x^{2}}{2}$ $v = x^2/2$

$u v - \int v d u$ $uv - intvdu$

$= x^2/2ln^3x - int x^cancel(2)/2 * (3ln^2x)/cancel(x)dx$

$= (x^2ln^3x)/2 - 3/2int xln^2xdx$

Repeat:
$u = ln^2x$
$du = (2lnx)/xdx$
$dv = xdx$
$v = x^2/2$

$= (x^2ln^3x)/2 - 3/2(int xln^2xdx)$

$= (x^2ln^3x)/2 - 3/2((x^2ln^2x)/2 - int x^cancel(2)/cancel(2) * (cancel(2)lnx)/cancel(x) dx)$

$= (x^2ln^3x)/2 - 3/2((x^2ln^2x)/2 - int xlnx dx)$

and repeat again:
$u = lnx$
$du = 1/xdx$
$dv = xdx$
$v = x^2/2$

$= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2(int xlnx dx)$

$= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2((x^2lnx)/2 - int x^cancel(2)/2*1/cancel(x)dx)$

$= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2((x^2lnx)/2 - int x/2dx)$

$= 1/2x^2ln^3x - 3/4x^2ln^2x + 3/4x^2lnx - 3/4int xdx$

$= 1/2x^2ln^3x - 3/4x^2ln^2x + 3/4x^2lnx - 3/8x^2$

$= color(blue)(1/8x^2(4ln^3x - 6ln^2x + 6lnx - 3) + C)$

Answer 2 · 2015-06-30T21:42:41

I found:
$x^2/2ln^3(x)-3/4x^2ln^2(x)+3/4x^2ln(x)-3/8x^2+c$

I started integrating by Substitution and the by Parts (three times):
enter image source here

How do you find the integral of x(ln x)^3 dxx(lnx)3dxx(ln x)^3 dx?