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

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

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

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

Therefore,

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

Add the constant of integration:

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

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