integral of (5x^2+20x+6)/(x^3+2x^2+x)

Lösung

\int \frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} d x

6 ln ∣ x ∣ - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + C

Schritte zur Lösung

\int \frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} d x

Ermittle den Partialbruch von

\frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} : \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{( x + 1 ) ^{2}}

= \int \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{( x + 1 ) ^{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 \frac{6}{x} d x - \int \frac{1}{x + 1} d x + \int \frac{9}{( x + 1 ) ^{2}} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

\int \frac{9}{( x + 1 ) ^{2}} d x = - \frac{9}{x + 1}

= 6 ln ∣ x ∣ - ln ∣ x + 1 ∣ - \frac{9}{x + 1}

Füge eine Konstante zur Lösung hinzu

= 6 ln ∣ x ∣ - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + C

y^{^{'}} - 4 y = 0.8

e x p an d (x - 1)^{2} (x - 2) (x - 4)

\int sinh (3 u) d u

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

n = 1 \sum \infty \frac{n ^{3}}{3 ^{n}}