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

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

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

\int_{0}^{x} ln (1 - x) d x

- x + x ln (- x + 1) - ln (- x + 1)

Solution steps

\int_{0}^{x} ln (1 - x) d x

Apply u-substitution

: \int_{1}^{1 - x} - ln (u) d u

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

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{1}^{1 - x}

= - [u ln (u) - \int 1 d u]_{1}^{1 - x}

\int 1 d u = u

= - [u ln (u) - u]_{1}^{1 - x}

Compute the boundaries:

x - x ln (- x + 1) + ln (- x + 1)

= - (x - x ln (- x + 1) + ln (- x + 1))

Simplify

= - x + x ln (- x + 1) - ln (- x + 1)

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