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

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Answer 1 · 2015-06-30T21:31:13

Let:
#u = ln^3x#
#du = (3ln^2x)/xdx#
#dv = xdx#
#v = x^2/2#

#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):
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