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

Lösung

\int \frac{5}{( x - 2 ) ( x ^{2} - 1 )} d x

5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

Schritte zur Lösung

\int \frac{5}{( x - 2 ) ( x ^{2} - 1 )} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{1}{( x - 2 ) ( x ^{2} - 1 )} d x

Wende U-Substitution an

= 5 \cdot \int \frac{1}{u ( u ^{2} + 4 u + 3 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u ^{2} + 4 u + 3 )} : \frac{1}{3 u} - \frac{1}{2 ( u + 1 )} + \frac{1}{6 ( u + 3 )}

= 5 \cdot \int \frac{1}{3 u} - \frac{1}{2 ( u + 1 )} + \frac{1}{6 ( u + 3 )} 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

= 5 (\int \frac{1}{3 u} d u - \int \frac{1}{2 ( u + 1 )} d u + \int \frac{1}{6 ( u + 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

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

\int \frac{1}{6 ( u + 3 )} d u = \frac{1}{6} ln ∣ u + 3 ∣

= 5 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{6} ln ∣ u + 3 ∣)

Setze in

u = x - 2

ein

= 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 2 + 1 ∣ + \frac{1}{6} ln ∣ x - 2 + 3 ∣)

Vereinfache

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

= 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

\int \frac{e ^{x^{\frac{1}{2}}}}{x ^{\frac{1}{2}}} d x

\frac{d}{d x} ((6 x)^{2 x})

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