integral of 5/(x(x^4+9))

Lösung

\int \frac{5}{x ( x ^{4} + 9 )} d x

\frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9 + C

Schritte zur Lösung

\int \frac{5}{x ( x ^{4} + 9 )} 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} + 9 )} d x

Wende U-Substitution an

= 5 \cdot \int \frac{1}{4 u ( u + 9 )} 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 + 9 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u + 9 )} : \frac{1}{9 u} - \frac{1}{9 ( u + 9 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{9 u} - \frac{1}{9 ( u + 9 )} 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}{9 u} d u - \int \frac{1}{9 ( u + 9 )} d u)

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u + 9 )} d u = \frac{1}{9} ln ∣ u + 9 ∣

= 5 \cdot \frac{1}{4} (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

Setze in

u = x^{4}

ein

= 5 \cdot \frac{1}{4} (\frac{1}{9} ln x^{4} - \frac{1}{9} ln x^{4} + 9)

Vereinfache

5 \cdot \frac{1}{4} (\frac{1}{9} ln x^{4} - \frac{1}{9} ln x^{4} + 9) : \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9

= \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9

Füge eine Konstante zur Lösung hinzu

= \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9 + C

\frac{d}{d x} (8 x + 1)

\int (x^{2} + 7) d x

\frac{d y}{d x} = \frac{x}{( 1 + 2 y )}

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

a re a y = x, y = - x, x = 1