How do you integrate #int ln(x+3)# by integration by parts method?

1 Answer
Morgan
Nov 26, 2016

First use #u#-substitution, then use integration by parts.

Explanation:

First, use #u# substitution, where #u=x+3#, #du=dx#, and rewrite as

#intln(u)du#

Using the integration by parts method, the integral can be rewritten in the form #uv-intvdu#.

To do this, you must choose some term in the original integral to set equal to #u# and one to set equal to #dv#. That which you set equal to #u# will be derived and that which you set equal to #dv# will be "anti-derived."

I would set #u=ln(u)# and #dv=1#. This gives #du=1/(u)du# and #v=u#. By the above form,

#u ln(u)-intu/udu#

#=>u ln(u)-int1du#

#=>u ln(u)-u+C#

#=>(x+3)ln(x+3)-(x+3)+C#

Hope that helps!

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