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