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

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 ∣

Add a constant to the solution

= - ln ∣ 1 - x ∣ + C

y^{^{'}} = sin (x)

x \to 1 lim (2 x^{2})

f (x) = \frac{( 2 - x ^{2} )}{2}, [- \frac{5}{4}, \frac{5}{4}]

k = 1 \sum \infty k^{3} e^{- k^{2}}

(x^{2} + y) d x + (x - e^{y}) d y = 0