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

Lösung

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

\frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int \frac{6 x}{x ^{3} - x ^{2} + x - 1} d x = 6 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

= \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 6 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

Vereinfache

\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + 6 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) : \frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x)

= \frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x)

Füge eine Konstante zur Lösung hinzu

= \frac{7}{2} ln ∣ x - 1 ∣ - \frac{5}{4} ln x^{2} + 1 + \frac{7}{2} arctan (x) + C

x \to 2 lim ((x + 1) \cdot (x + 3))

\frac{\partial}{\partial x} (x e^{- x} (a + b x + c x^{2}))

\frac{\partial}{\partial x} (x^{3} y^{2} - 2 x y^{3})

d er i v a t i v e (x^{2} - 3) (x^{3} - 2 x)

\int_{3}^{9} 1 + x^{4} + 2 x^{2} d x