integral of (x-3)/(x^2+3x-4)

Lösung

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

- \frac{2}{5} ln ∣ x - 1 ∣ + \frac{7}{5} ln ∣ x + 4 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

\frac{x - 3}{x ^{2} + 3 x - 4} : \frac{x}{x ^{2} + 3 x - 4} - \frac{3}{x ^{2} + 3 x - 4}

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

\int \frac{x}{x ^{2} + 3 x - 4} d x = \frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣

\int \frac{3}{x ^{2} + 3 x - 4} d x = 3 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

= \frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣ - 3 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

Vereinfache

\frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣ - 3 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣) : - \frac{2}{5} ln ∣ x - 1 ∣ + \frac{7}{5} ln ∣ x + 4 ∣

= - \frac{2}{5} ln ∣ x - 1 ∣ + \frac{7}{5} ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{2}{5} ln ∣ x - 1 ∣ + \frac{7}{5} ln ∣ x + 4 ∣ + C

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