Integral of $$$- x^{23} + x + 1$$$
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Find $$$\int \left(- x^{23} + x + 1\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(- x^{23} + x + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{x d x} - \int{x^{23} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\int{x d x} - \int{x^{23} d x} + {\color{red}{\int{1 d x}}} = \int{x d x} - \int{x^{23} d x} + {\color{red}{x}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$x - \int{x^{23} d x} + {\color{red}{\int{x d x}}}=x - \int{x^{23} d x} + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=x - \int{x^{23} d x} + {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=23$$$:
$$\frac{x^{2}}{2} + x - {\color{red}{\int{x^{23} d x}}}=\frac{x^{2}}{2} + x - {\color{red}{\frac{x^{1 + 23}}{1 + 23}}}=\frac{x^{2}}{2} + x - {\color{red}{\left(\frac{x^{24}}{24}\right)}}$$
Therefore,
$$\int{\left(- x^{23} + x + 1\right)d x} = - \frac{x^{24}}{24} + \frac{x^{2}}{2} + x$$
Simplify:
$$\int{\left(- x^{23} + x + 1\right)d x} = \frac{x \left(- x^{23} + 12 x + 24\right)}{24}$$
Add the constant of integration:
$$\int{\left(- x^{23} + x + 1\right)d x} = \frac{x \left(- x^{23} + 12 x + 24\right)}{24}+C$$
Answer
$$$\int \left(- x^{23} + x + 1\right)\, dx = \frac{x \left(- x^{23} + 12 x + 24\right)}{24} + C$$$A