What is the integral of ((3w^2)/2-2/(3w^2)) ?

The integral of ((3w^2)/2-2/(3w^2)) is (w^3)/2+2/(3w)+C

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integral of ((3w^2)/2-2/(3w^2))

\int (\frac{3 w ^{2}}{2} - \frac{2}{3 w ^{2}}) d w

\frac{w ^{3}}{2} + \frac{2}{3 w} + C

Solution steps

\int \frac{3 w ^{2}}{2} - \frac{2}{3 w ^{2}} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 w ^{2}}{2} d w - \int \frac{2}{3 w ^{2}} d w

\int \frac{3 w ^{2}}{2} d w = \frac{w ^{3}}{2}

\int \frac{2}{3 w ^{2}} d w = - \frac{2}{3 w}

= \frac{w ^{3}}{2} - (- \frac{2}{3 w})

Simplify

= \frac{w ^{3}}{2} + \frac{2}{3 w}

Add a constant to the solution

= \frac{w ^{3}}{2} + \frac{2}{3 w} + C

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What is the integral of ((3w^2)/2-2/(3w^2)) ?
The integral of ((3w^2)/2-2/(3w^2)) is (w^3)/2+2/(3w)+C