What is the integral from 0 to x of 1/(1-x) ?

The integral from 0 to x of 1/(1-x) is -ln|-x+1|

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integral from 0 to x of 1/(1-x)

\int_{0}^{x} \frac{1}{1 - x} d x

- ln ∣ - x + 1 ∣

Solution steps

\int_{0}^{x} \frac{1}{1 - x} d x

Apply u-substitution

: \int_{1}^{1 - x} - \frac{1}{u} d u

= \int_{1}^{1 - x} - \frac{1}{u} d u

Take the constant out:

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

= - \int_{1}^{1 - x} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{1}^{1 - x}

Compute the boundaries:

ln ∣ - x + 1 ∣

= - ln ∣ - x + 1 ∣

\frac{d}{d θ} (\frac{1}{θ})

\frac{\partial}{\partial x} (6 e^{x^{3} y})

\int 26 d x

f (x) = 3 x^{3} - x^{2} + 8, (2, 28)

\frac{d}{d x} (8 - x^{2})

What is the integral from 0 to x of 1/(1-x) ?
The integral from 0 to x of 1/(1-x) is -ln|-x+1|