integral of (t^4+9)/(t^4+9t^2)

Lösung

\int \frac{t ^{4} + 9}{t ^{4} + 9 t ^{2}} d t

t - \frac{1}{t} - \frac{10}{3} arctan (\frac{t}{3}) + C

Schritte zur Lösung

\int \frac{t ^{4} + 9}{t ^{4} + 9 t ^{2}} d t

Ermittle den Partialbruch von

\frac{t ^{4} + 9}{t ^{4} + 9 t ^{2}} : 1 + \frac{1}{t ^{2}} - \frac{10}{t ^{2} + 9}

= \int 1 + \frac{1}{t ^{2}} - \frac{10}{t ^{2} + 9} 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

= \int 1 d t + \int \frac{1}{t ^{2}} d t - \int \frac{10}{t ^{2} + 9} d t

\int 1 d t = t

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

\int \frac{10}{t ^{2} + 9} d t = \frac{10}{3} arctan (\frac{t}{3})

= t - \frac{1}{t} - \frac{10}{3} arctan (\frac{t}{3})

Füge eine Konstante zur Lösung hinzu

= t - \frac{1}{t} - \frac{10}{3} arctan (\frac{t}{3}) + C