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

1 Answer
Nov 26, 2016

First use uu-substitution, then use integration by parts.

Explanation:

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

intln(u)duln(u)du

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

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

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

u ln(u)-intu/uduuln(u)uudu

=>u ln(u)-int1duuln(u)1du

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

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

Hope that helps!