integral of (x^2+5x-4)/(x^3-x)

Lösung

\int \frac{x ^{2} + 5 x - 4}{x ^{3} - x} d x

\frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{x ^{2} + 5 x - 4}{x ^{3} - x} d x

Multipliziere aus

\frac{x ^{2} + 5 x - 4}{x ^{3} - x} : \frac{x ^{2}}{x ^{3} - x} + \frac{5 x}{x ^{3} - x} - \frac{4}{x ^{3} - 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{x ^{2}}{x ^{3} - x} d x + \int \frac{5 x}{x ^{3} - x} d x - \int \frac{4}{x ^{3} - x} d x

\int \frac{x ^{2}}{x ^{3} - x} d x = \frac{1}{2} ln x^{2} - 1

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

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

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

Vereinfache

\frac{1}{2} ln x^{2} - 1 - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) - 4 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) : \frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

= \frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

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

\int cos (x + 5) d x

\int e^{- \frac{x ^{2}}{5}} d x

\int (\frac{7}{1 - x ^{2}} - \frac{8}{x}) d x

\int 2 x + \frac{x}{2} d x