# What is the net area between f(x)=ln(x+1) in x in[1,2]  and the x-axis?

Mar 21, 2017

$3 \ln 3 - 2 \ln 2 - 1$

#### Explanation:

${\int}_{1}^{2} \ln \left(x + 1\right) \mathrm{dx}$

First change variables:
Let $\textcolor{red}{w = x + 1}$
$\mathrm{dw} = \mathrm{dx}$

Next, solve $\int \ln \left(\textcolor{red}{w}\right) \mathrm{dw}$ using integration by parts.

$\int \left(\ln w\right) \mathrm{dw}$
$\textcolor{b l u e}{u = \ln w \text{ } v = w}$
$\textcolor{b l u e}{\mathrm{du} = \frac{1}{w} \mathrm{dw} \text{ } \mathrm{dv} = 1}$

$\int \left(\ln w\right) \mathrm{dw} = w \ln w - \int \left(w \cdot \frac{1}{w}\right) \mathrm{dw}$
$= w \ln w - \int \left(1\right) \mathrm{dw}$
$= \textcolor{red}{w} \ln \textcolor{red}{w} - \textcolor{red}{w}$
$= \left(x + 1\right) \ln \left(x + 1\right) - \left(x + 1\right)$
$= \left(x + 1\right) \ln \left(x + 1\right) - x - 1$

Now, go back to the definite integral:
${\int}_{1}^{2} \ln \left(x + 1\right) \mathrm{dx}$

$= {\left[\left(x + 1\right) \ln \left(x + 1\right) - x - 1\right]}_{1}^{2}$

$= \left[\left(2 + 1\right) \ln \left(2 + 1\right) - 2 - 1\right] - \left[\left(1 + 1\right) \ln \left(1 + 1\right) - 1 - 1\right]$

$= 3 \ln 3 - 3 - \left(2 \ln 2 - 2\right)$

$= 3 \ln 3 - 3 - 2 \ln 2 + 2$

$= 3 \ln 3 - 2 \ln 2 - 1$