# How do you integrate int ln(x+1) by integration by parts method?

Oct 19, 2016

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

and then putting
$u = \ln \left(x + 1\right)$ & $\frac{\mathrm{dv}}{\mathrm{dx}} = 1$

we end up with the result:

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

#### Explanation:

Integration by parts formula:

$\int u \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \mathrm{dx} = u v - \int v \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \mathrm{dx}$

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

with intergation by parts it is usual to put $u = \ln f \left(x\right)$

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

Let$\frac{\mathrm{dv}}{\mathrm{dx}} = 1 \implies v = x$

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

on spliting the improper fraction in the integral we have:

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

integrating:

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

which simplifies to:

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