integral of (2x^3+x-1)/(x^2-5x)

Lösung

\int \frac{2 x ^{3} + x - 1}{x ^{2} - 5 x} d x

x^{2} + 10 x + \frac{1}{5} ln ∣ x ∣ + \frac{254}{5} ln ∣ x - 5 ∣ - 75 + C

Schritte zur Lösung

\int \frac{2 x ^{3} + x - 1}{x ^{2} - 5 x} d x

Multipliziere aus

\frac{2 x ^{3} + x - 1}{x ^{2} - 5 x} : \frac{2 x ^{3}}{x ^{2} - 5 x} + \frac{x}{x ^{2} - 5 x} - \frac{1}{x ^{2} - 5 x}

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 ^{3}}{x ^{2} - 5 x} d x + \int \frac{x}{x ^{2} - 5 x} d x - \int \frac{1}{x ^{2} - 5 x} d x

\int \frac{2 x ^{3}}{x ^{2} - 5 x} d x = x^{2} + 10 x + 50 ln ∣ x - 5 ∣ - 75

\int \frac{x}{x ^{2} - 5 x} d x = ln ∣ x - 5 ∣

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

= x^{2} + 10 x + 50 ln ∣ x - 5 ∣ - 75 + ln ∣ x - 5 ∣ - (- \frac{1}{5} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 5 ∣)

Vereinfache

x^{2} + 10 x + 50 ln ∣ x - 5 ∣ - 75 + ln ∣ x - 5 ∣ - (- \frac{1}{5} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 5 ∣) : x^{2} + 10 x + \frac{1}{5} ln ∣ x ∣ + \frac{254}{5} ln ∣ x - 5 ∣ - 75

= x^{2} + 10 x + \frac{1}{5} ln ∣ x ∣ + \frac{254}{5} ln ∣ x - 5 ∣ - 75

Füge eine Konstante zur Lösung hinzu

= x^{2} + 10 x + \frac{1}{5} ln ∣ x ∣ + \frac{254}{5} ln ∣ x - 5 ∣ - 75 + C

in t e g r a l e^{5 x}

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