integral of 1/(t^4+16)

Lösung

\int \frac{1}{t ^{4} + 16} d t

\frac{1}{32 2} (ln t^{2} + 22 t + 4 + 2 arctan (\frac{t + 2}{2})) + \frac{1}{32 2} (- ln t^{2} - 22 t + 4 + 2 arctan (\frac{t - 2}{2})) + C

Schritte zur Lösung

\int \frac{1}{t ^{4} + 16} d t

Ermittle den Partialbruch von

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

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

\int \frac{t + 2 2}{16 2 ( t ^{2} + 2 2 t + 4 )} d t = \frac{1}{32 2} (ln t^{2} + 22 t + 4 + 2 arctan (\frac{t + 2}{2}))

\int \frac{- t + 2 2}{16 2 ( t ^{2} - 2 2 t + 4 )} d t = \frac{1}{32 2} (- ln t^{2} - 22 t + 4 + 2 arctan (\frac{t - 2}{2}))

= \frac{1}{32 2} (ln t^{2} + 22 t + 4 + 2 arctan (\frac{t + 2}{2})) + \frac{1}{32 2} (- ln t^{2} - 22 t + 4 + 2 arctan (\frac{t - 2}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{32 2} (ln t^{2} + 22 t + 4 + 2 arctan (\frac{t + 2}{2})) + \frac{1}{32 2} (- ln t^{2} - 22 t + 4 + 2 arctan (\frac{t - 2}{2})) + C

x \to - 3 + lim (\frac{x + 2}{x + 3})

x \to 3 lim (\frac{x + 2}{x + 3})

\int (2 x^{3} + 5 cos (x) + e^{x}) d x

\int x (1 - x) d x

\int cos (\frac{1}{4} x) d x