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

Nov 26, 2016

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

Explanation:

First, use $u$ substitution, where $u = x + 3$, $\mathrm{du} = \mathrm{dx}$, and rewrite as

$\int \ln \left(u\right) \mathrm{du}$

Using the integration by parts method, the integral can be rewritten in the form $u v - \int v \mathrm{du}$.

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

I would set $u = \ln \left(u\right)$ and $\mathrm{dv} = 1$. This gives $\mathrm{du} = \frac{1}{u} \mathrm{du}$ and $v = u$. By the above form,

$u \ln \left(u\right) - \int \frac{u}{u} \mathrm{du}$

$\implies u \ln \left(u\right) - \int 1 \mathrm{du}$

$\implies u \ln \left(u\right) - u + C$

$\implies \left(x + 3\right) \ln \left(x + 3\right) - \left(x + 3\right) + C$

Hope that helps!