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

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

   2. Let $u = \sqrt{x + 1}$.

      Then let $du = \frac{dx}{2 \sqrt{x + 1}}$ and substitute $2 du$:

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

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

         $$\int \frac{1}{u^{2}}\, du = 2 \int \frac{1}{u^{2}}\, du$$

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

            $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

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

      Now substitute $u$ back in:

      $$- \frac{2}{\sqrt{x + 1}}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Let $u = \sqrt{x + 1}$.

      Then let $du = \frac{dx}{2 \sqrt{x + 1}}$ and substitute $2 du$:

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

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

         $$\int \frac{1}{u^{2}}\, du = 2 \int \frac{1}{u^{2}}\, du$$

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

            $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

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

      Now substitute $u$ back in:

      $$- \frac{2}{\sqrt{x + 1}}$$

2. Add the constant of integration:

   $$- \frac{2}{\sqrt{x + 1}}+ \mathrm{constant}$$

The answer is:

$$- \frac{2}{\sqrt{x + 1}}+ \mathrm{constant}$$