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

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

Expand the expression:

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

Integrate term by term:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

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

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

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

Therefore,

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

Add the constant of integration:

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

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