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Beliebt Rechnen >

integral of 8/(t^2(t^2-4))

Lösung

\int \frac{8}{t ^{2} ( t ^{2} - 4 )} d t

Lösung

8 (\frac{1}{4 t} - \frac{1}{16} ln ∣ t + 2 ∣ + \frac{1}{16} ln ∣ t - 2 ∣) + C

Schritte zur Lösung

\int \frac{8}{t ^{2} ( t ^{2} - 4 )} d t

Entferne die Konstante:

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

= 8 \cdot \int \frac{1}{t ^{2} ( t ^{2} - 4 )} d t

Ermittle den Partialbruch von

\frac{1}{t ^{2} ( t ^{2} - 4 )} : - \frac{1}{4 t ^{2}} - \frac{1}{16 ( t + 2 )} + \frac{1}{16 ( t - 2 )}

= 8 \cdot \int - \frac{1}{4 t ^{2}} - \frac{1}{16 ( t + 2 )} + \frac{1}{16 ( t - 2 )} d t

Wende die Summenregel an:

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

= 8 (- \int \frac{1}{4 t ^{2}} d t - \int \frac{1}{16 ( t + 2 )} d t + \int \frac{1}{16 ( t - 2 )} d t)

\int \frac{1}{4 t ^{2}} d t = - \frac{1}{4 t}

\int \frac{1}{16 ( t + 2 )} d t = \frac{1}{16} ln ∣ t + 2 ∣

\int \frac{1}{16 ( t - 2 )} d t = \frac{1}{16} ln ∣ t - 2 ∣

= 8 (- (- \frac{1}{4 t}) - \frac{1}{16} ln ∣ t + 2 ∣ + \frac{1}{16} ln ∣ t - 2 ∣)

Vereinfache

= 8 (\frac{1}{4 t} - \frac{1}{16} ln ∣ t + 2 ∣ + \frac{1}{16} ln ∣ t - 2 ∣)

Füge eine Konstante zur Lösung hinzu

= 8 (\frac{1}{4 t} - \frac{1}{16} ln ∣ t + 2 ∣ + \frac{1}{16} ln ∣ t - 2 ∣) + C

Graph

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