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

The integral of ln(1-x) is -ln(1-x)+xln(1-x)+1-x+C

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integral of ln(1-x)

\int ln (1 - x) d x

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

Solution steps

\int ln (1 - x) d x

Apply u-substitution

: \int - ln (u) d u

= \int - ln (u) d u

Take the constant out:

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

= - \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= - (u ln (u) - \int 1 d u)

\int 1 d u = u

= - (u ln (u) - u)

Substitute back

u = 1 - x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d t} = 2 (t + 1) y, y (0) = 1

\frac{d}{d x} (e^{x} - x e^{x})

\int_{1}^{2} 3 r^{4} ln (r) d r

x \to 2 lim (6 x + 13)

\int \frac{3 x}{4 - 2 x ^{2}} d x

What is the integral of ln(1-x) ?
The integral of ln(1-x) is -ln(1-x)+xln(1-x)+1-x+C