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

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

Expand the expression:

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

Integrate term by term:

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

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

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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