integral of (4x^2+3x+19)/((x+1)(x^2+4))

Lösung

\int \frac{4 x ^{2} + 3 x + 19}{( x + 1 ) ( x ^{2} + 4 )} d x

4 ln ∣ x + 1 ∣ + \frac{3}{2} arctan (\frac{x}{2}) + C

Schritte zur Lösung

\int \frac{4 x ^{2} + 3 x + 19}{( x + 1 ) ( x ^{2} + 4 )} d x

Ermittle den Partialbruch von

\frac{4 x ^{2} + 3 x + 19}{( x + 1 ) ( x ^{2} + 4 )} : \frac{4}{x + 1} + \frac{3}{x ^{2} + 4}

= \int \frac{4}{x + 1} + \frac{3}{x ^{2} + 4} d x

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{4}{x + 1} d x + \int \frac{3}{x ^{2} + 4} d x

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

\int \frac{3}{x ^{2} + 4} d x = \frac{3}{2} arctan (\frac{x}{2})

= 4 ln ∣ x + 1 ∣ + \frac{3}{2} arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= 4 ln ∣ x + 1 ∣ + \frac{3}{2} arctan (\frac{x}{2}) + C

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