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

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

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integral of 1/z-x/(y^2)

\int \frac{1}{z} - \frac{x}{y ^{2}} d y

\frac{y}{z} + \frac{x}{y} + C

Solution steps

\int \frac{1}{z} - \frac{x}{y ^{2}} d y

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

\int \frac{1}{z} d y = \frac{y}{z}

\int \frac{x}{y ^{2}} d y = - \frac{x}{y}

= \frac{y}{z} - (- \frac{x}{y})

Simplify

= \frac{y}{z} + \frac{x}{y}

Add a constant to the solution

= \frac{y}{z} + \frac{x}{y} + C

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