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 ∣))

Vereinfache

= 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

d er i v a t i v e f (x) = \frac{4 x}{x ^{2} + 1}

\frac{d}{d x} ((3 x - 5)^{2})

x \to 0 lim (\frac{5}{x ^{2} + x} - \frac{5}{x})

t a y l or \frac{1}{2 - x}, 0

\frac{d}{d x} (\frac{4 x ^{2}}{x ^{2} - 1})