Integral of $$$\ln\left(x + 1\right)$$$

Let $$$u=x + 1$$$.

Then $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\ln{\left(x + 1 \right)} d x}}} = {\color{red}{\int{\ln{\left(u \right)} d u}}}$$

For the integral $$$\int{\ln{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{h} \operatorname{dv} = \operatorname{h}\operatorname{v} - \int \operatorname{v} \operatorname{dh}$$$.

Let $$$\operatorname{h}=\ln{\left(u \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{dh}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

The integral becomes

$${\color{red}{\int{\ln{\left(u \right)} d u}}}={\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}={\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$u \ln{\left(u \right)} - {\color{red}{\int{1 d u}}} = u \ln{\left(u \right)} - {\color{red}{u}}$$

Recall that $$$u=x + 1$$$:

$$- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - {\color{red}{\left(x + 1\right)}} + {\color{red}{\left(x + 1\right)}} \ln{\left({\color{red}{\left(x + 1\right)}} \right)}$$

Therefore,

$$\int{\ln{\left(x + 1 \right)} d x} = - x + \left(x + 1\right) \ln{\left(x + 1 \right)} - 1$$

Simplify:

$$\int{\ln{\left(x + 1 \right)} d x} = \left(x + 1\right) \left(\ln{\left(x + 1 \right)} - 1\right)$$

Add the constant of integration:

$$\int{\ln{\left(x + 1 \right)} d x} = \left(x + 1\right) \left(\ln{\left(x + 1 \right)} - 1\right)+C$$

Answer: $$$\int{\ln{\left(x + 1 \right)} d x}=\left(x + 1\right) \left(\ln{\left(x + 1 \right)} - 1\right)+C$$$