What is the integral of (1/(z^3)-3/(z^2)) ?

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

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integral of (1/(z^3)-3/(z^2))

\int (\frac{1}{z ^{3}} - \frac{3}{z ^{2}}) d z

- \frac{1}{2 z ^{2}} + \frac{3}{z} + C

Solution steps

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

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

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

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

= - \frac{1}{2 z ^{2}} - (- \frac{3}{z})

Simplify

= - \frac{1}{2 z ^{2}} + \frac{3}{z}

Add a constant to the solution

= - \frac{1}{2 z ^{2}} + \frac{3}{z} + C

\frac{\partial}{\partial y} (x^{2} + 4 y^{2})

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

\int \frac{4 - x ^{2}}{x ^{2}} d x

\int_{0}^{11} x 11 - x d x

x \to 3 lim (5 x^{3})

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