integral of (4x^2-x+1)/(x^3-x^2+x-1)

Lösung

\int \frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} d x

2 ln ∣ x - 1 ∣ + ln x^{2} + 1 + arctan (x) + C

Schritte zur Lösung

\int \frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} d x

Multipliziere aus

\frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} : \frac{4 x ^{2}}{x ^{3} - x ^{2} + x - 1} - \frac{x}{x ^{3} - x ^{2} + x - 1} + \frac{1}{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{4 x ^{2}}{x ^{3} - x ^{2} + x - 1} d x - \int \frac{x}{x ^{3} - x ^{2} + x - 1} d x + \int \frac{1}{x ^{3} - x ^{2} + x - 1} d x

\int \frac{4 x ^{2}}{x ^{3} - x ^{2} + x - 1} d x = 4 (\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{1}{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))

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

= 2 ln ∣ x - 1 ∣ + ln x^{2} + 1 + arctan (x) + C