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

Lösung

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

- 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3 + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

Vereinfache

\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ - (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣ : - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3

= - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3

Füge eine Konstante zur Lösung hinzu

= - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3 + C

in t e g r a l \frac{1}{2 ^{x}}

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