integral of 1/(x+1)x

Lösung

\int \frac{1}{x + 1} x d x

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

Schritte zur Lösung

\int \frac{1}{x + 1} x d x

Wende die partielle Integration an

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

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