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

The integral of (y-1)/(y^2) is ln|y|+1/y+C

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

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

ln ∣ y ∣ + \frac{1}{y} + C

Solution steps

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

Expand

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

= \int \frac{1}{y} - \frac{1}{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}{y} d y - \int \frac{1}{y ^{2}} d y

\int \frac{1}{y} d y = ln ∣ y ∣

\int \frac{1}{y ^{2}} d y = - \frac{1}{y}

= ln ∣ y ∣ - (- \frac{1}{y})

Simplify

= ln ∣ y ∣ + \frac{1}{y}

Add a constant to the solution

= ln ∣ y ∣ + \frac{1}{y} + C

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What is the integral of (y-1)/(y^2) ?
The integral of (y-1)/(y^2) is ln|y|+1/y+C