integral of 1/(1+t^3)

Lösung

\int \frac{1}{1 + t ^{3}} d t

\frac{1}{3} ln ∣ t + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 t ^{2} - 4 t}{3} + 23 arctan (\frac{2 t - 1}{3})) + C

Schritte zur Lösung

\int \frac{1}{1 + t ^{3}} d t

Ermittle den Partialbruch von

\frac{1}{1 + t ^{3}} : \frac{1}{3 ( t + 1 )} + \frac{- t + 2}{3 ( t ^{2} - t + 1 )}

= \int \frac{1}{3 ( t + 1 )} + \frac{- t + 2}{3 ( t ^{2} - t + 1 )} 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{1}{3 ( t + 1 )} d t + \int \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t

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

\int \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 t ^{2} - 4 t}{3} + 23 arctan (\frac{2 t - 1}{3}))

= \frac{1}{3} ln ∣ t + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 t ^{2} - 4 t}{3} + 23 arctan (\frac{2 t - 1}{3}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} ln ∣ t + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 t ^{2} - 4 t}{3} + 23 arctan (\frac{2 t - 1}{3})) + C

\int \frac{1}{7 ^{x}} d x

t an g e n t f (x) = - x^{2} + 7 x

\int_{- 1}^{1} x^{3} - 6 x d x

t an g e n t x^{4} + 5 x^{2} - x

d er i v a t i v e h (w) = \frac{w ^{2} - 1}{w ^{2} + 1}