Integral of $$$\frac{1}{u^{2}}$$$
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Your Input
Find $$$\int \frac{1}{u^{2}}\, du$$$.
Solution
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:
$${\color{red}{\int{\frac{1}{u^{2}} d u}}}={\color{red}{\int{u^{-2} d u}}}={\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}={\color{red}{\left(- u^{-1}\right)}}={\color{red}{\left(- \frac{1}{u}\right)}}$$
Therefore,
$$\int{\frac{1}{u^{2}} d u} = - \frac{1}{u}$$
Add the constant of integration:
$$\int{\frac{1}{u^{2}} d u} = - \frac{1}{u}+C$$
Answer
$$$\int \frac{1}{u^{2}}\, du = - \frac{1}{u} + C$$$A
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