What is the integral of u^{-3} 1/2 ?

The integral of u^{-3} 1/2 is -1/(4u^2)+C

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integral of u^{-3} 1/2

\int u^{- 3} \frac{1}{2} d u

- \frac{1}{4 u ^{2}} + C

Solution steps

\int u^{- 3} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{- 3} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{2} (- \frac{1}{2 u ^{2}})

Simplify

\frac{1}{2} (- \frac{1}{2 u ^{2}}) : - \frac{1}{4 u ^{2}}

= - \frac{1}{4 u ^{2}}

Add a constant to the solution

= - \frac{1}{4 u ^{2}} + C

\frac{d y}{d x} = x e^{- x}

\int_{0}^{7} \frac{5}{3} x d x

- \frac{1}{s ^{2} + ( 5 ) ^{2}}

\int_{- 3}^{0} 9 - x^{2} d x

\frac{3 s - 1}{( s ^{2} + 2 s + 5 )}