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

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

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integral of-(ln(x))/(y^2)

\int - \frac{ln ( x )}{y ^{2}} d y

\frac{ln ( x )}{y} + C

Solution steps

\int - \frac{ln ( x )}{y ^{2}} d y

Take the constant out:

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

= - ln (x) \cdot \int \frac{1}{y ^{2}} d y

Apply the Power Rule

: - \frac{1}{y}

= - ln (x) (- \frac{1}{y})

Simplify

- ln (x) (- \frac{1}{y}) : \frac{ln ( x )}{y}

= \frac{ln ( x )}{y}

Add a constant to the solution

= \frac{ln ( x )}{y} + C

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