Integral of $$$\frac{1}{x^{2} + 4}$$$

Find $$$\int \frac{1}{x^{2} + 4}\, dx$$$.

Let $$$u=\frac{x}{2}$$$.

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

The integral can be rewritten as

$$\color{red}{\int{\frac{1}{x^{2} + 4} d x}} = \color{red}{\int{\frac{1}{2 \left(u^{2} + 1\right)} d u}}$$

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

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

The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

Recall that $$$u=\frac{x}{2}$$$:

$$\frac{\operatorname{atan}{\left(\color{red}{u} \right)}}{2} = \frac{\operatorname{atan}{\left(\color{red}{\left(\frac{x}{2}\right)} \right)}}{2}$$

Therefore,

$$\int{\frac{1}{x^{2} + 4} d x} = \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$$

Add the constant of integration:

$$\int{\frac{1}{x^{2} + 4} d x} = \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}+C$$

Answer: $$$\int{\frac{1}{x^{2} + 4} d x}=\frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}+C$$$