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

Lösung

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

\frac{1}{2} x^{2} ln (\frac{x - 1}{x + 1}) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x + C

Schritte zur Lösung

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} x^{2} ln (\frac{x - 1}{x + 1}) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x + C

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