How do you integrate #int ln x^3 dx # using integration by parts?

1 Answer
mason m
Apr 19, 2016

#intln(x^3)dx=3xlnx-3x+C#

Explanation:

First, simplify the function using the rule #ln(a^b)=b*lna#; this gives the simplified integral

#intln(x^3)dx=int3lnxdx=3intlnxdx#

The question then becomes, how do we integrate #intlnxdx# by parts?

Integration by parts takes the form

#intudv=uv-intvdu#

So, we have to determine what to let #u# and #dv# equal within #intlnxdx#. Our best bet is to let #u=lnx# and simply have #dv=1dx#, since there's not much else we can do.

Find the values of #du# and #v#, by differentiating and integrating respectively.

#ul(u=lnx)" "=>" "(du)/dx=1/x" "=>" "ul(du=1/xdx)#

#ul(dv=1dx)" "=>" "intdv=intdx" "=>" "ul(v=x#

Plugging these into the integration by parts formula, not forgetting the #3# tagging along multiplicatively, we have:

#3intlnxdx=3(xlnx-intx(1/x)dx)#

#=3xlnx-3intdx#

#=3xlnx-3x+C#

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