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

Lösung

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

\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{3}{5} ln ∣ x + 4 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{2 x}{( x + 4 ) ( x ^{2} - 1 )} d x = 2 (- \frac{4}{15} ln ∣ x + 4 ∣ + \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣)

\int \frac{1}{( x + 4 ) ( x ^{2} - 1 )} d x = \frac{1}{15} ln ∣ x + 4 ∣ - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣

= 2 (- \frac{4}{15} ln ∣ x + 4 ∣ + \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣) - (\frac{1}{15} ln ∣ x + 4 ∣ - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣)

Vereinfache

2 (- \frac{4}{15} ln ∣ x + 4 ∣ + \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣) - (\frac{1}{15} ln ∣ x + 4 ∣ - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣) : \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{3}{5} ln ∣ x + 4 ∣

= \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{3}{5} ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{3}{5} ln ∣ x + 4 ∣ + C

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