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

The integral of 1/(yIn^2y) is -1/(In^2y)+C

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integral of 1/(yIn^2y)

\int \frac{1}{y I n ^{2} y} d y

- \frac{1}{I n ^{2} y} + C

Solution steps

\int \frac{1}{y I n ^{2} y} d y

Take the constant out:

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

= \frac{1}{I n ^{2}} \cdot \int \frac{1}{yy} d y

yy = y^{2}

= \frac{1}{I n ^{2}} \cdot \int \frac{1}{y ^{2}} d y

Apply the Power Rule

: - \frac{1}{y}

= \frac{1}{I n ^{2}} (- \frac{1}{y})

Simplify

\frac{1}{I n ^{2}} (- \frac{1}{y}) : - \frac{1}{I n ^{2} y}

= - \frac{1}{I n ^{2} y}

Add a constant to the solution

= - \frac{1}{I n ^{2} y} + C

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