integral of (3x)/((x+1)(x-2))

Lösung

\int \frac{3 x}{( x + 1 ) ( x - 2 )} d x

ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{3 x}{( x + 1 ) ( x - 2 )} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{( x + 1 ) ( x - 2 )} d x

Ermittle den Partialbruch von

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

= 3 \cdot \int \frac{1}{3 ( x + 1 )} + \frac{2}{3 ( x - 2 )} 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

= 3 (\int \frac{1}{3 ( x + 1 )} d x + \int \frac{2}{3 ( x - 2 )} d x)

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

\int \frac{2}{3 ( x - 2 )} d x = \frac{2}{3} ln ∣ x - 2 ∣

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

Vereinfache

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

= ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ x + 1 ∣ + 2 ln ∣ x - 2 ∣ + C

n \to \infty lim (\frac{1}{5})

y^{^{'}} = \frac{2 y + x}{x}

n = 0 \sum \infty ln (2)

\int \frac{3 x}{x ^{2}} d x

x \to 4 lim (\frac{2}{( x - 4 ) ^{2}})