# How do you integrate ln(x+1)?

Aug 28, 2015

Notice how you can write this as:

$\int 1 \cdot \ln \left(x + 1\right) \mathrm{dx}$

To integrate this, you can do Integration by Parts.

Let:
$u = \ln \left(x + 1\right)$
$\mathrm{du} = \frac{1}{x + 1} \mathrm{dx}$
$\mathrm{dv} = 1 \mathrm{dx}$
$v = x$

$u v - \int v \mathrm{du}$

$= x \ln \left(x + 1\right) - \int \frac{x}{x + 1} \mathrm{dx}$

$= x \ln \left(x + 1\right) - \int \frac{x + 1 - 1}{x + 1} \mathrm{dx}$

$= x \ln \left(x + 1\right) - \int 1 - \frac{1}{x + 1} \mathrm{dx}$

$= x \ln \left(x + 1\right) - \left(x - \ln \left(x + 1\right)\right)$

$= x \ln \left(x + 1\right) + \ln \left(x + 1\right) - x$

$= \textcolor{b l u e}{\left(x + 1\right) \ln \left(x + 1\right) - x + C}$