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

The integral of (-2)/(z^3) is 1/(z^2)+C

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

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

\frac{1}{z ^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

Simplify

- 2 (- \frac{1}{2 z ^{2}}) : \frac{1}{z ^{2}}

= \frac{1}{z ^{2}}

Add a constant to the solution

= \frac{1}{z ^{2}} + C

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + e^{- x}

\frac{d}{d x} (\frac{sin ( x )}{1 + sin ( x )})

\int_{1}^{\infty} (\frac{1}{x ^{3}}) d x

l a pl a ce t r an s f or m 4 t^{2} cos (2 t) + 5 t e^{- 7 t}

\frac{d}{d x} (4 x^{9} + 2 x^{5})