What is the integral of u^{-3}(9-u^3+u^6) ?

The integral of u^{-3}(9-u^3+u^6) is -9/(2u^2)-u+(u^4)/4+C

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integral of u^{-3}(9-u^3+u^6)

\int u^{- 3} (9 - u^{3} + u^{6}) d u

- \frac{9}{2 u ^{2}} - u + \frac{u ^{4}}{4} + C

Solution steps

\int u^{- 3} (9 - u^{3} + u^{6}) d u

Expand

u^{- 3} (9 - u^{3} + u^{6}) : \frac{9}{u ^{3}} - 1 + u^{3}

= \int \frac{9}{u ^{3}} - 1 + u^{3} d u

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{9}{u ^{3}} d u - \int 1 d u + \int u^{3} d u

\int \frac{9}{u ^{3}} d u = - \frac{9}{2 u ^{2}}

\int 1 d u = u

\int u^{3} d u = \frac{u ^{4}}{4}

= - \frac{9}{2 u ^{2}} - u + \frac{u ^{4}}{4}

Add a constant to the solution

= - \frac{9}{2 u ^{2}} - u + \frac{u ^{4}}{4} + C

What is the integral of u^{-3}(9-u^3+u^6) ?
The integral of u^{-3}(9-u^3+u^6) is -9/(2u^2)-u+(u^4)/4+C