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

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

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

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

- x - ln ∣ - x + 1 ∣

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{- u + 1}{u} : - 1 + \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= - (- (- x) + ln ∣ - x + 1 ∣)

Simplify

= - x - ln ∣ - x + 1 ∣

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