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

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

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integral of (-1)/(1-x)

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

ln ∣ 1 - x ∣ + C

Solution steps

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

Take the constant out:

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

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

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 - x

= - (- ln ∣ 1 - x ∣)

Simplify

= ln ∣ 1 - x ∣

Add a constant to the solution

= ln ∣ 1 - x ∣ + C

\int_{1}^{3} (- x^{2} + 4 x - 3) d x

x \to (0) lim (\frac{ln ( x )}{x})

x \to - 3 lim (2 x - 3 a x + 6 a)

x \to 1 + lim (\frac{∣ x - 1 ∣}{x - 1})

\frac{\partial}{\partial x} (\frac{- x}{y ^{2} + x ^{2}})