Integral of $$$- x^{2} + \frac{1}{a^{2}}$$$ with respect to $$$x$$$
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Find $$$\int \left(- x^{2} + \frac{1}{a^{2}}\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(- x^{2} + \frac{1}{a^{2}}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{a^{2}} d x} - \int{x^{2} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=\frac{1}{a^{2}}$$$:
$$- \int{x^{2} d x} + {\color{red}{\int{\frac{1}{a^{2}} d x}}} = - \int{x^{2} d x} + {\color{red}{\frac{x}{a^{2}}}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$- {\color{red}{\int{x^{2} d x}}} + \frac{x}{a^{2}}=- {\color{red}{\frac{x^{1 + 2}}{1 + 2}}} + \frac{x}{a^{2}}=- {\color{red}{\left(\frac{x^{3}}{3}\right)}} + \frac{x}{a^{2}}$$
Therefore,
$$\int{\left(- x^{2} + \frac{1}{a^{2}}\right)d x} = - \frac{x^{3}}{3} + \frac{x}{a^{2}}$$
Add the constant of integration:
$$\int{\left(- x^{2} + \frac{1}{a^{2}}\right)d x} = - \frac{x^{3}}{3} + \frac{x}{a^{2}}+C$$
Answer
$$$\int \left(- x^{2} + \frac{1}{a^{2}}\right)\, dx = \left(- \frac{x^{3}}{3} + \frac{x}{a^{2}}\right) + C$$$A