What is the integral of-1/y ?

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

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integral of-1/y

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

(x - 9) (x - 3)^{2}

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

y = x^{5 s i n (x)}

y^{^{'}} = \frac{( x ^{3} y ^{3} + 2 x ^{3} )}{( y ^{2} )}

\frac{2}{( s - 2 ) ^{4}}