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{sin ( x ) - x}{x ^{3}} d x

y^{^{'}} = 4 (y^{2} + 1)

x \to \infty lim (\frac{1}{x} - 5)

x \to 3 lim (\frac{4}{x ^{2} + 2 x + 3})

\frac{d}{d x} (14 - \frac{1}{2} x^{- 3})

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