integral of 1/(x(4+x^6))

Lösung

\int \frac{1}{x ( 4 + x ^{6} )} d x

\frac{1}{6} (\frac{3}{2} ln (x) - \frac{1}{4} ln x^{6} + 4) + C

Schritte zur Lösung

\int \frac{1}{x ( 4 + x ^{6} )} d x

Wende U-Substitution an

= \int \frac{1}{6 u ( 4 + u )} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int \frac{1}{u ( 4 + u )} d u

Ermittle den Partialbruch von

\frac{1}{u ( 4 + u )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= \frac{1}{6} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( 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

= \frac{1}{6} (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

= \frac{1}{6} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

Setze in

u = x^{6}

ein

= \frac{1}{6} (\frac{1}{4} ln x^{6} - \frac{1}{4} ln x^{6} + 4)

\frac{1}{4} ln x^{6} = \frac{3}{2} ln (x)

= \frac{1}{6} (\frac{3}{2} ln (x) - \frac{1}{4} ln x^{6} + 4)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} (\frac{3}{2} ln (x) - \frac{1}{4} ln x^{6} + 4) + C

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