integral of ln((1-x)/(1+x))

Lösung

\int ln (\frac{1 - x}{1 + x}) d x

x ln (\frac{1 - x}{1 + x}) - ln (∣ x + 1 ∣ ∣ - x + 1 ∣) + C

Schritte zur Lösung

\int ln (\frac{1 - x}{1 + x}) d x

Wende die partielle Integration an

= x ln (\frac{1 - x}{1 + x}) - \int - \frac{2 x}{( - x + 1 ) ( x + 1 )} d x

\int - \frac{2 x}{( - x + 1 ) ( x + 1 )} d x = ln (∣ x + 1 ∣ ∣ - x + 1 ∣)

= x ln (\frac{1 - x}{1 + x}) - ln (∣ x + 1 ∣ ∣ - x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= x ln (\frac{1 - x}{1 + x}) - ln (∣ x + 1 ∣ ∣ - x + 1 ∣) + C

\frac{d ^{2}}{d x ^{2}} (3 e^{- x^{2}})

\frac{d}{d x} (\frac{20 sin ( tw )}{tw})

t an g e n t y = - 5 - 8 x^{2}, (- 3, - 77)

x \frac{d y}{d x} + y = 11 x

\int e^{- \frac{x}{3}} d x