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Beliebt Rechnen >

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

Lösung

\int \frac{1}{( x ^{2} - 1 ) ( x - 3 )} d x

Lösung

\frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{( x ^{2} - 1 ) ( x - 3 )} d x

Wende U-Substitution an

= \int \frac{1}{u ( u ^{2} + 6 u + 8 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u ^{2} + 6 u + 8 )} : \frac{1}{8 u} - \frac{1}{4 ( u + 2 )} + \frac{1}{8 ( u + 4 )}

= \int \frac{1}{8 u} - \frac{1}{4 ( u + 2 )} + \frac{1}{8 ( u + 4 )} d u

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{1}{8 u} d u - \int \frac{1}{4 ( u + 2 )} d u + \int \frac{1}{8 ( u + 4 )} d u

\int \frac{1}{8 u} d u = \frac{1}{8} ln ∣ u ∣

\int \frac{1}{4 ( u + 2 )} d u = \frac{1}{4} ln ∣ u + 2 ∣

\int \frac{1}{8 ( u + 4 )} d u = \frac{1}{8} ln ∣ u + 4 ∣

= \frac{1}{8} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 2 ∣ + \frac{1}{8} ln ∣ u + 4 ∣

Setze in

u = x - 3

ein

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 3 + 2 ∣ + \frac{1}{8} ln ∣ x - 3 + 4 ∣

Vereinfache

\frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 3 + 2 ∣ + \frac{1}{8} ln ∣ x - 3 + 4 ∣ : \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣ + C

Graph

Beliebte Beispiele

integral of 2sqrt(x-2)

\int 2 x - 2 d x

y^'+y/t =t^2y^4

y^{^{'}} + \frac{y}{t} = t^{2} y^{4}

limit as x approaches+4+of 7/(x-4)

x \to + 4 + lim (\frac{7}{x - 4})

integral of (x^6-x^3+16)/(x^4+4x^2)

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

t(dy)/(dt)=t^3+15t^3y

t \frac{d y}{d t} = t^{3} + 15 t^{3} y