Integral of ln(1+x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x + 1$.

      Then let $du = dx$ and substitute $du$:

      $$\int \log{\left(u \right)}\, du$$

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$.

         Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$.

         To find $v{\left(u \right)}$:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int 1\, du = u$$

         Now evaluate the sub-integral.

      2. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, du = u$$

      Now substitute $u$ back in:

      $$- x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1$$

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \log{\left(x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}$.

      To find $v{\left(x \right)}$:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      Now evaluate the sub-integral.

   2. Rewrite the integrand:

      $$\frac{x}{x + 1} = 1 - \frac{1}{x + 1}$$

   3. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      1. The integral of a constant times a function is the constant times the integral of the function:

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

         1. Let $u = x + 1$.

            Then let $du = dx$ and substitute $du$:

            $$\int \frac{1}{u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(x + 1 \right)}$$

         So, the result is: $- \log{\left(x + 1 \right)}$

      The result is: $x - \log{\left(x + 1 \right)}$

2. Add the constant of integration:

   $$- x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1+ \mathrm{constant}$$

The answer is:

$$- x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1+ \mathrm{constant}$$