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

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

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

\int \frac{1}{y ln ^{2} ( y )} d y

- \frac{1}{ln ( y )} + C

Solution steps

\int \frac{1}{y ln ^{2} ( y )} d y

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = ln (y)

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

Add a constant to the solution

= - \frac{1}{ln ( y )} + C

\int \frac{( x ^{2} + 9 )}{x ^{2} + 2 x + 3} d x

y^{^{''}} + y^{^{'}} = e^{x}

\int \frac{1}{x ^{- 1}} d x

\frac{d y}{d t} = y (10 - y)

\frac{\partial}{\partial x} (cos (θ))

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