integral of (48)/(x^4-10x^2+9)

Lösung

\int \frac{48}{x ^{4} - 10 x ^{2} + 9} d x

3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣ + C

Schritte zur Lösung

\int \frac{48}{x ^{4} - 10 x ^{2} + 9} d x

Entferne die Konstante:

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

= 48 \cdot \int \frac{1}{x ^{4} - 10 x ^{2} + 9} d x

Ermittle den Partialbruch von

\frac{1}{x ^{4} - 10 x ^{2} + 9} : \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} - \frac{1}{48 ( x + 3 )} + \frac{1}{48 ( x - 3 )}

= 48 \cdot \int \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} - \frac{1}{48 ( x + 3 )} + \frac{1}{48 ( x - 3 )} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 48 (\int \frac{1}{16 ( x + 1 )} d x - \int \frac{1}{16 ( x - 1 )} d x - \int \frac{1}{48 ( x + 3 )} d x + \int \frac{1}{48 ( x - 3 )} d x)

\int \frac{1}{16 ( x + 1 )} d x = \frac{1}{16} ln ∣ x + 1 ∣

\int \frac{1}{16 ( x - 1 )} d x = \frac{1}{16} ln ∣ x - 1 ∣

\int \frac{1}{48 ( x + 3 )} d x = \frac{1}{48} ln ∣ x + 3 ∣

\int \frac{1}{48 ( x - 3 )} d x = \frac{1}{48} ln ∣ x - 3 ∣

= 48 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{48} ln ∣ x + 3 ∣ + \frac{1}{48} ln ∣ x - 3 ∣)

Vereinfache

48 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{48} ln ∣ x + 3 ∣ + \frac{1}{48} ln ∣ x - 3 ∣) : 3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣

= 3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= 3 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣ + C

\frac{d}{d x} ((3 x) (x + 3))

y^{^{''}} + 25 y = 2 sec (5 t)

x \to 0 lim (5^{x})

n = 6 \sum \infty e^{3 - 4 n}

x \to \infty lim (4 x - x^{2})