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

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

Let $$$u=x^{2}$$$.

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

So,

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

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 $$$u=x^{2}$$$:

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

Therefore,

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

Add the constant of integration:

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

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