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

Lösung

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

\frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16} + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{x ^{2}}{2 x - 1} d x = \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3)

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

= \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣)

Vereinfache

\frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) : \frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16}

= \frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16} + C

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