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

Lösung

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

\frac{1}{96} (2 (2 x - 1)^{3} + 9 (2 x - 1)^{2} + 18 (2 x - 1) + 6 ln ∣ 2 x - 1 ∣) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{x ^{3}}{2 x - 1} d x = \frac{1}{96} (2 (2 x - 1)^{3} + 9 (2 x - 1)^{2} + 18 (2 x - 1) + 6 ln ∣ 2 x - 1 ∣)

\int \frac{x}{2 x - 1} d x = \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣)

= \frac{1}{96} (2 (2 x - 1)^{3} + 9 (2 x - 1)^{2} + 18 (2 x - 1) + 6 ln ∣ 2 x - 1 ∣) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{96} (2 (2 x - 1)^{3} + 9 (2 x - 1)^{2} + 18 (2 x - 1) + 6 ln ∣ 2 x - 1 ∣) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) + C