What is the integral of 1/(2v^{-2)} ?

The integral of 1/(2v^{-2)} is (v^3)/6+C

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integral of 1/(2v^{-2)}

\int \frac{1}{2 v ^{- 2}} d v

\frac{v ^{3}}{6} + C

Solution steps

\int \frac{1}{2 v ^{- 2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{v ^{- 2}} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{2} \cdot \frac{v ^{3}}{3}

Simplify

\frac{1}{2} \cdot \frac{v ^{3}}{3} : \frac{v ^{3}}{6}

= \frac{v ^{3}}{6}

Add a constant to the solution

= \frac{v ^{3}}{6} + C

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