How do you integrate #(ln x / 3)^3#?

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Answer 1 · 2016-04-20T16:40:55

#int(lnx/3)^3dx=(xln^3x)/27-(xln^2x)/9+(2xlnx)/9-(2x)/9+C#

Note that #(lnx/3)^3=ln^3x/27#. From this, we see that

#int(lnx/3)^3dx=1/27intln^3xdx#

#intudv=uv-intvdu#

We let

#u=ln^3x" "=>" "du=(3ln^2x)/xdx#

#dv=(1)dx" "=>" "v=x#

This gives us:

#1/27intln^3xdx=1/27xln^3x-1/27int3ln^2xdx#

#=1/27xln^3x-1/9intln^2xdx#

Integrate #intln^2xdx# similarly (using by parts again):

#u=ln^2x" "=>" "du=(2lnx)/xdx#

#dv=(1)dx" "=>" "v=x#

Thus,

#intln^2x=xln^2x-int2lnxdx#

Combining this and multiplying by #-1/9#, we see that:

#1/27intln^3xdx=(xln^3x)/27-(xln^2x)/9+2/9intlnxdx#

Use integration by parts one last time:

#u=lnx" "=>" "du=1/xdx#

#dv=(1)dx" "=>" "v=x#

Thus,

#intlnxdx=xlnx-intdx=xlnx-x#

Hence,

#1/27intln^3xdx=(xln^3x)/27-(xln^2x)/9+(2xlnx)/9-(2x)/9+C#

Don't forget the constant of integration!