What is the integral of u^{-3}(4-u^3+u^6) ?

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

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integral of u^{-3}(4-u^3+u^6)

\int u^{- 3} (4 - u^{3} + u^{6}) d u

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

Solution steps

\int u^{- 3} (4 - u^{3} + u^{6}) d u

Expand

u^{- 3} (4 - u^{3} + u^{6}) : \frac{4}{u ^{3}} - 1 + u^{3}

= \int \frac{4}{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{4}{u ^{3}} d u - \int 1 d u + \int u^{3} d u

\int \frac{4}{u ^{3}} d u = - \frac{2}{u ^{2}}

\int 1 d u = u

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

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

Add a constant to the solution

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

e^{x} (x^{3} - 3 x^{2} + 6 x - 6)

\frac{\partial}{\partial z} (\frac{x - z}{y + z})

cos (x) \frac{π}{4}

x \to \infty lim (tan (x))

\frac{d}{d x} (\frac{3}{( x + 1 ) ^{2}})

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