Integral of $$$d_{\nu}^{2} x + x$$$ with respect to $$$x$$$

Find $$$\int \left(d_{\nu}^{2} x + x\right)\, dx$$$.

Integrate term by term:

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

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

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

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

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

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

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

Therefore,

$$\int{\left(d_{\nu}^{2} x + x\right)d x} = \frac{d_{\nu}^{2} x^{2}}{2} + \frac{x^{2}}{2}$$

Simplify:

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

Add the constant of integration:

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

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