integral of (5x+1)/((2x+1)(x-1))x

Lösung

\int \frac{5 x + 1}{( 2 x + 1 ) ( x - 1 )} x d x

\frac{5}{2} x - \frac{1}{4} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + \frac{1}{4} - 2 + C

Schritte zur Lösung

\int \frac{5 x + 1}{( 2 x + 1 ) ( x - 1 )} x d x

Wende die partielle Integration an

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

\int \frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ d x = \frac{1}{2} x ln ∣ 2 x + 1 ∣ - \frac{5}{2} x + 2 x ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ 2 x + 1 ∣ - 2 ln ∣ x - 1 ∣ - \frac{1}{4} + 2

= x (\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣) - (\frac{1}{2} x ln ∣ 2 x + 1 ∣ - \frac{5}{2} x + 2 x ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ 2 x + 1 ∣ - 2 ln ∣ x - 1 ∣ - \frac{1}{4} + 2)

Vereinfache

x (\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣) - (\frac{1}{2} x ln ∣ 2 x + 1 ∣ - \frac{5}{2} x + 2 x ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ 2 x + 1 ∣ - 2 ln ∣ x - 1 ∣ - \frac{1}{4} + 2) : \frac{5}{2} x - \frac{1}{4} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + \frac{1}{4} - 2

= \frac{5}{2} x - \frac{1}{4} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + \frac{1}{4} - 2

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} x - \frac{1}{4} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + \frac{1}{4} - 2 + C