Integral of (ln(2x+2))/(x+1) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \log{\left(2 x + 2 \right)}$.

      Then let $du = \frac{2 dx}{2 x + 2}$ and substitute $du$:

      $$\int u\, du$$

      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int u\, du = \frac{u^{2}}{2}$$

      Now substitute $u$ back in:

      $$\frac{\log{\left(2 x + 2 \right)}^{2}}{2}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{\log{\left(2 x + 2 \right)}}{x + 1} = \frac{\log{\left(x + 1 \right)} + \log{\left(2 \right)}}{x + 1}$$

   2. Let $u = x + 1$.

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

      $$\int \frac{\log{\left(u \right)} + \log{\left(2 \right)}}{u}\, du$$

      1. Let $u = \frac{1}{u}$.

         Then let $du = - \frac{du}{u^{2}}$ and substitute $du$:

         $$\int \frac{- \log{\left(\frac{1}{u} \right)} - \log{\left(2 \right)}}{u}\, du$$

         1. Let $u = - \log{\left(\frac{1}{u} \right)} - \log{\left(2 \right)}$.

            Then let $du = \frac{du}{u}$ and substitute $du$:

            $$\int u\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u\, du = \frac{u^{2}}{2}$$

            Now substitute $u$ back in:

            $$\frac{\left(- \log{\left(\frac{1}{u} \right)} - \log{\left(2 \right)}\right)^{2}}{2}$$

         Now substitute $u$ back in:

         $$\frac{\left(- \log{\left(u \right)} - \log{\left(2 \right)}\right)^{2}}{2}$$

      Now substitute $u$ back in:

      $$\frac{\left(- \log{\left(x + 1 \right)} - \log{\left(2 \right)}\right)^{2}}{2}$$

2. Now simplify:

   $$\frac{\log{\left(2 x + 2 \right)}^{2}}{2}$$

3. Add the constant of integration:

   $$\frac{\log{\left(2 x + 2 \right)}^{2}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(2 x + 2 \right)}^{2}}{2}+ \mathrm{constant}$$