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

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

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integral of 1/(y-yln(y))

\int \frac{1}{y - y ln ( y )} d y

- ln ∣ 1 - ln (y) ∣ + C

Solution steps

\int \frac{1}{y - y ln ( y )} d y

Apply u-substitution

: \int \frac{1}{1 - u} d u

= \int \frac{1}{1 - u} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

= - ln ∣ 1 - ln (y) ∣

Add a constant to the solution

= - ln ∣ 1 - ln (y) ∣ + C

4 - 3 x

f (x) = ln (sin (x) + 4)

\frac{d}{d x} (1000 (sin (x)) (e^{- 2 x}))

\int \frac{cos ( 9 x )}{1 + sin ^{2} ( 9 x )} d x

\int \frac{x}{x ^{2} + y} d x

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