Integral of $$$\frac{1}{\sqrt{x} + 1}$$$

Find $$$\int \frac{1}{\sqrt{x} + 1}\, dx$$$.

Let $$$u=\sqrt{x}$$$.

Then $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

The integral becomes

$${\color{red}{\int{\frac{1}{\sqrt{x} + 1} d x}}} = {\color{red}{\int{\frac{2 u}{u + 1} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = \frac{u}{u + 1}$$$:

$${\color{red}{\int{\frac{2 u}{u + 1} d u}}} = {\color{red}{\left(2 \int{\frac{u}{u + 1} d u}\right)}}$$

Rewrite and split the fraction:

$$2 {\color{red}{\int{\frac{u}{u + 1} d u}}} = 2 {\color{red}{\int{\left(1 - \frac{1}{u + 1}\right)d u}}}$$

Integrate term by term:

$$2 {\color{red}{\int{\left(1 - \frac{1}{u + 1}\right)d u}}} = 2 {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u + 1} d u}\right)}}$$

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

$$- 2 \int{\frac{1}{u + 1} d u} + 2 {\color{red}{\int{1 d u}}} = - 2 \int{\frac{1}{u + 1} d u} + 2 {\color{red}{u}}$$

Let $$$v=u + 1$$$.

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

Therefore,

$$2 u - 2 {\color{red}{\int{\frac{1}{u + 1} d u}}} = 2 u - 2 {\color{red}{\int{\frac{1}{v} d v}}}$$

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$2 u - 2 {\color{red}{\int{\frac{1}{v} d v}}} = 2 u - 2 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$

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

$$2 u - 2 \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = 2 u - 2 \ln{\left(\left|{{\color{red}{\left(u + 1\right)}}}\right| \right)}$$

Recall that $$$u=\sqrt{x}$$$:

$$- 2 \ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)} + 2 {\color{red}{u}} = - 2 \ln{\left(\left|{1 + {\color{red}{\sqrt{x}}}}\right| \right)} + 2 {\color{red}{\sqrt{x}}}$$

Therefore,

$$\int{\frac{1}{\sqrt{x} + 1} d x} = 2 \sqrt{x} - 2 \ln{\left(\left|{\sqrt{x} + 1}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{x} + 1} d x} = 2 \sqrt{x} - 2 \ln{\left(\left|{\sqrt{x} + 1}\right| \right)}+C$$

$$$\int \frac{1}{\sqrt{x} + 1}\, dx = \left(2 \sqrt{x} - 2 \ln\left(\left|{\sqrt{x} + 1}\right|\right)\right) + C$$$A