integral of (6x^2+x-1)/(3x^3-9x^2+x-3)

Lösung

\int \frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} d x

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

Schritte zur Lösung

\int \frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} d x

Ermittle den Partialbruch von

\frac{6 x ^{2} + x - 1}{3 x ^{3} - 9 x ^{2} + x - 3} : \frac{2}{x - 3} + \frac{1}{3 x ^{2} + 1}

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e tan (e^{6 t}) + e^{t a n (6 t)}

\int - e^{2 t} + t e^{t} d t

\int_{0}^{1} x 6 - x^{2} d x

\frac{d}{d x} (sin (x) + \frac{3}{4} cot (x))

\frac{\partial}{\partial x} (cos (x y^{2}))