integral ln(1-z)

Lösung

integral

ln (1 - z)

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

Schritte zur Lösung

\int ln (1 - z) d z

Wende U-Substitution an

= \int - ln (u) d u

Entferne die Konstante:

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

= - \int ln (u) d u

Wende die partielle Integration an

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

\int 1 d u = u

= - (u ln (u) - u)

Setze in

u = 1 - z

ein

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

Vereinfache

- ((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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{1}{3 ^{x}}

in t e g r a l e^{- x - 3}

in t e g r a l r^{3}

in t e g r a l \frac{x}{( x + 3 )}

in t e g r a l sin (t) cos^{2} (t)