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

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

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integral of 1/((1-y))

integral

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

- ln ∣ 1 - y ∣ + C

Solution steps

\int \frac{1}{1 - y} d y

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 - y

= - ln ∣ 1 - y ∣

Add a constant to the solution

= - ln ∣ 1 - y ∣ + C

(\frac{ln ( x )}{x})^{2}

\frac{x ^{3}}{( 1 - x ^{4} )}

sin^{2} (\frac{x}{2})

\frac{( 1 + x )}{( 1 + x ^{2} )}

y^{- 1}