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

Lösung

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

ln ∣ x - 1 ∣ - arctan (x) + C

Schritte zur Lösung

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

Multipliziere aus

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

= ln ∣ x - 1 ∣ - arctan (x)

Füge eine Konstante zur Lösung hinzu

= ln ∣ x - 1 ∣ - arctan (x) + C

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

t an g e n t f (x) = \frac{1}{4 x ^{\frac{3}{2}}}, at x = 4

\frac{d y}{d x} = 6 x^{2} y

n = 1 \sum \infty n e^{- 8 n}

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