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

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

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

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

- ln ∣ - x + 1 ∣

Solution steps

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

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 ∣

\int_{0}^{4} (4 x - 3 x) d x

\int_{2}^{4} \frac{2 x}{x ^{2} - 1} d x

\int_{0}^{\infty} e

\int_{- 3}^{7} \frac{2 x + 6}{5} d x

\int_{- 5}^{5} - x^{2} + 25 d x

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