integral of (6(1+s))/(s(s^2+7s+6))

Lösung

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

ln ∣ s ∣ - ln ∣ s + 6 ∣ + C

Schritte zur Lösung

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

Entferne die Konstante:

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

= 6 \cdot \int \frac{1 + s}{s ( s ^{2} + 7 s + 6 )} d s

Ermittle den Partialbruch von

\frac{1 + s}{s ( s ^{2} + 7 s + 6 )} : \frac{1}{6 s} - \frac{1}{6 ( s + 6 )}

= 6 \cdot \int \frac{1}{6 s} - \frac{1}{6 ( s + 6 )} d s

Wende die Summenregel an:

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

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

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

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

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

Vereinfache

6 (\frac{1}{6} ln ∣ s ∣ - \frac{1}{6} ln ∣ s + 6 ∣) : ln ∣ s ∣ - ln ∣ s + 6 ∣

= ln ∣ s ∣ - ln ∣ s + 6 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ s ∣ - ln ∣ s + 6 ∣ + C

\int 1 - z^{2}

h \to 0 lim (\frac{1}{h})

x \to 0 lim (\frac{x}{sin ( 5 x )})

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

\int_{0}^{π} x^{2} cos (n x) d x