integral of (x^2+2x+5)/(x+4)

Lösung

\int \frac{x ^{2} + 2 x + 5}{x + 4} d x

- 6 x + 13 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 24 + C

Schritte zur Lösung

\int \frac{x ^{2} + 2 x + 5}{x + 4} d x

Multipliziere aus

\frac{x ^{2} + 2 x + 5}{x + 4} : \frac{x ^{2}}{x + 4} + \frac{2 x}{x + 4} + \frac{5}{x + 4}

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 + 4} d x + \int \frac{2 x}{x + 4} d x + \int \frac{5}{x + 4} d x

\int \frac{x ^{2}}{x + 4} d x = \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣

\int \frac{2 x}{x + 4} d x = 2 (x + 4 - 4 ln ∣ x + 4 ∣)

\int \frac{5}{x + 4} d x = 5 ln ∣ x + 4 ∣

= \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ + 2 (x + 4 - 4 ln ∣ x + 4 ∣) + 5 ln ∣ x + 4 ∣

Vereinfache

\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ + 2 (x + 4 - 4 ln ∣ x + 4 ∣) + 5 ln ∣ x + 4 ∣ : - 6 x + 13 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 24

= - 6 x + 13 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 24

Füge eine Konstante zur Lösung hinzu

= - 6 x + 13 ln ∣ x + 4 ∣ + \frac{1}{2} (x + 4)^{2} - 24 + C

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