Integral of $$$x + 1$$$
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Solution
Integrate term by term:
$${\color{red}{\int{\left(x + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{x d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\int{x d x} + {\color{red}{\int{1 d x}}} = \int{x d x} + {\color{red}{x}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$x + {\color{red}{\int{x d x}}}=x + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=x + {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Therefore,
$$\int{\left(x + 1\right)d x} = \frac{x^{2}}{2} + x$$
Simplify:
$$\int{\left(x + 1\right)d x} = \frac{x \left(x + 2\right)}{2}$$
Add the constant of integration:
$$\int{\left(x + 1\right)d x} = \frac{x \left(x + 2\right)}{2}+C$$
Answer
$$$\int \left(x + 1\right)\, dx = \frac{x \left(x + 2\right)}{2} + C$$$A