integral of 5/(x(x^4+6))

Lösung

\int \frac{5}{x ( x ^{4} + 6 )} d x

\frac{5}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6) + C

Schritte zur Lösung

\int \frac{5}{x ( x ^{4} + 6 )} 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 ( x ^{4} + 6 )} d x

Wende U-Substitution an

= 5 \cdot \int \frac{1}{4 u ( u + 6 )} d u

Entferne die Konstante:

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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 6 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u + 6 )} : \frac{1}{6 u} - \frac{1}{6 ( u + 6 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{6 u} - \frac{1}{6 ( u + 6 )} 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 \cdot \frac{1}{4} (\int \frac{1}{6 u} d u - \int \frac{1}{6 ( u + 6 )} d u)

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

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

= 5 \cdot \frac{1}{4} (\frac{1}{6} ln ∣ u ∣ - \frac{1}{6} ln ∣ u + 6 ∣)

Setze in

u = x^{4}

ein

= 5 \cdot \frac{1}{4} (\frac{1}{6} ln x^{4} - \frac{1}{6} ln x^{4} + 6)

Vereinfache

5 \cdot \frac{1}{4} (\frac{1}{6} ln x^{4} - \frac{1}{6} ln x^{4} + 6) : \frac{5}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6)

= \frac{5}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6)

Füge eine Konstante zur Lösung hinzu

= \frac{5}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6) + C

x \to 2 lim (\frac{2 - x}{x})

\int_{- 2}^{2} \frac{3}{( x + 3 ) ^{4}} d x

\int y (x^{2} + y^{2}) d y

f (x) = 6 x

d er i v a t i v e f (x) = 3 e^{(2 x^{2})}