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

Take the constant out:

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

= - \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))

Integral of a constant:

\int a d x = a x

= - (- (u ln (u) - 1 \cdot u))

Substitute back

u = 1 - x

= - (- ((1 - x) ln (1 - x) - 1 \cdot (1 - x)))

Simplify

- (- ((1 - x) ln (1 - x) - 1 \cdot (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

x + y = 2

\int cos^{3} (\frac{x}{18}) d x

\int (5 x + 1) d x

y = \frac{5 x}{x + 2}, (3, 3)

\frac{\partial}{\partial x} (1 (x - y) e^{y + 2 x^{2}})

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