integral of (x-1)/(x^2+2x+1)

Lösung

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

ln ∣ x + 1 ∣ + \frac{2}{x + 1} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{x - 1}{x ^{2} + 2 x + 1} : \frac{x}{x ^{2} + 2 x + 1} - \frac{1}{x ^{2} + 2 x + 1}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Vereinfache

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

= ln ∣ x + 1 ∣ + \frac{2}{x + 1}

Füge eine Konstante zur Lösung hinzu

= ln ∣ x + 1 ∣ + \frac{2}{x + 1} + C

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