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

Lösung

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

4 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 4 + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= \frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣ + x - 1 + ln ∣ x - 1 ∣ - 4 ln ∣ x - 1 ∣

Vereinfache

\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣ + x - 1 + ln ∣ x - 1 ∣ - 4 ln ∣ x - 1 ∣ : 4 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 4

= 4 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 4

Füge eine Konstante zur Lösung hinzu

= 4 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 4 + C

\int (x^{2} - 4 x) e^{x} d x

\int (5 x^{4} - 3) d x

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

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\int \frac{- 5}{3 x} d x