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

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

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integral of (-1)/y

integral

\frac{- 1}{y}

- ln ∣ y ∣ + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= - ln ∣ y ∣

Add a constant to the solution

= - ln ∣ y ∣ + C

f (x) = \frac{1}{( 1 - x ) ^{\frac{1}{2}}}

cos^{2} (\frac{x}{2})

\frac{1}{( x + a ) ^{2}}

\frac{1}{( 2 n - 1 ) ^{2}}

\frac{2 x}{x + 2}