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

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \frac{dx}{x + 1}$ and substitute $du$:

      $$\int \sqrt{u}\, du$$

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

         $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

      Now substitute $u$ back in:

      $$\frac{2 \log{\left(x + 1 \right)}^{\frac{3}{2}}}{3}$$

   Method #2

   1. Let $u = x + 1$.

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

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

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

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

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

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

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

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

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

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

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

                  $$\int \sqrt{u}\, du = - \int \sqrt{u}\, du$$

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

                     $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

                  So, the result is: $- \frac{2 u^{\frac{3}{2}}}{3}$

               Now substitute $u$ back in:

               $$- \frac{2 \log{\left(\frac{1}{u} \right)}^{\frac{3}{2}}}{3}$$

            So, the result is: $\frac{2 \log{\left(\frac{1}{u} \right)}^{\frac{3}{2}}}{3}$

         Now substitute $u$ back in:

         $$\frac{2 \log{\left(u \right)}^{\frac{3}{2}}}{3}$$

      Now substitute $u$ back in:

      $$\frac{2 \log{\left(x + 1 \right)}^{\frac{3}{2}}}{3}$$

2. Now simplify:

   $$\frac{2 \log{\left(x + 1 \right)}^{\frac{3}{2}}}{3}$$

3. Add the constant of integration:

   $$\frac{2 \log{\left(x + 1 \right)}^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{2 \log{\left(x + 1 \right)}^{\frac{3}{2}}}{3}+ \mathrm{constant}$$