integral of (2x^3+x^2-7x+7)/(x^2+x-2)

Lösung

\int \frac{2 x ^{3} + x ^{2} - 7 x + 7}{x ^{2} + x - 2} d x

x^{2} - x + ln ∣ x - 1 ∣ - 3 ln ∣ x + 2 ∣ + C

Schritte zur Lösung

\int \frac{2 x ^{3} + x ^{2} - 7 x + 7}{x ^{2} + x - 2} d x

Ermittle den Partialbruch von

\frac{2 x ^{3} + x ^{2} - 7 x + 7}{x ^{2} + x - 2} : 2 x - 1 + \frac{1}{x - 1} - \frac{3}{x + 2}

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

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

\int 2 x d x = x^{2}

\int 1 d x = x

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

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

= x^{2} - x + ln ∣ x - 1 ∣ - 3 ln ∣ x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= x^{2} - x + ln ∣ x - 1 ∣ - 3 ln ∣ x + 2 ∣ + C

\frac{\partial}{\partial y} (x y e^{4 y})

\int_{1}^{4} 1 + \frac{9}{4} x d x

\frac{d}{d x} (\frac{x}{1 - 2 x ^{2}})

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

d er i v a t i v e 2 e^{- x}