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

The integral of ln(1-z) is -ln(1-z)+zln(1-z)+1-z+C

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integral of ln(1-z)

integral

ln (1 - z)

- ln (1 - z) + z ln (1 - z) + 1 - z + C

Solution steps

\int ln (1 - z) d z

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 - z

= - ((1 - z) ln (1 - z) - (1 - z))

Simplify

- ((1 - z) ln (1 - z) - (1 - z)) : - ln (1 - z) + z ln (1 - z) + 1 - z

= - ln (1 - z) + z ln (1 - z) + 1 - z

Add a constant to the solution

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

\frac{2 x}{( 1 + x ^{2} )}

\frac{1}{3 ^{x}}

\frac{e ^{x}}{( 2 + e ^{x} )}

e^{- x - 3}

r^{3}

What is the integral of ln(1-z) ?
The integral of ln(1-z) is -ln(1-z)+zln(1-z)+1-z+C