Integral of $$$- u^{2} + \frac{1}{a^{2}}$$$ with respect to $$$u$$$

The calculator will find the integral/antiderivative of $$$- u^{2} + \frac{1}{a^{2}}$$$ with respect to $$$u$$$, with steps shown.

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Find $$$\int \left(- u^{2} + \frac{1}{a^{2}}\right)\, du$$$.

Solution

Integrate term by term:

$${\color{red}{\int{\left(- u^{2} + \frac{1}{a^{2}}\right)d u}}} = {\color{red}{\left(\int{\frac{1}{a^{2}} d u} - \int{u^{2} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=\frac{1}{a^{2}}$$$:

$$- \int{u^{2} d u} + {\color{red}{\int{\frac{1}{a^{2}} d u}}} = - \int{u^{2} d u} + {\color{red}{\frac{u}{a^{2}}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$- {\color{red}{\int{u^{2} d u}}} + \frac{u}{a^{2}}=- {\color{red}{\frac{u^{1 + 2}}{1 + 2}}} + \frac{u}{a^{2}}=- {\color{red}{\left(\frac{u^{3}}{3}\right)}} + \frac{u}{a^{2}}$$

Therefore,

$$\int{\left(- u^{2} + \frac{1}{a^{2}}\right)d u} = - \frac{u^{3}}{3} + \frac{u}{a^{2}}$$

Add the constant of integration:

$$\int{\left(- u^{2} + \frac{1}{a^{2}}\right)d u} = - \frac{u^{3}}{3} + \frac{u}{a^{2}}+C$$

Answer

$$$\int \left(- u^{2} + \frac{1}{a^{2}}\right)\, du = \left(- \frac{u^{3}}{3} + \frac{u}{a^{2}}\right) + C$$$A