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

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

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

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

- ln ∣ - t + 1 ∣

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{1}^{1 - t}

Compute the boundaries:

ln ∣ - t + 1 ∣

= - ln ∣ - t + 1 ∣

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 8} d x

\int_{0}^{1} \frac{6 x ^{2}}{1 - x ^{3}} d x

\int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 y ^{2}} d y

\int_{1}^{\infty} \frac{4}{3 x - 1} d x

\int_{0}^{\frac{π}{3}} 2 cos (x) - 1 d x

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