integral of (t^6+1)/(t^6+t^3)

Lösung

\int \frac{t ^{6} + 1}{t ^{6} + t ^{3}} d t

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

Schritte zur Lösung

\int \frac{t ^{6} + 1}{t ^{6} + t ^{3}} d t

Ermittle den Partialbruch von

\frac{t ^{6} + 1}{t ^{6} + t ^{3}} : 1 + \frac{1}{t ^{3}} - \frac{2}{3 ( t + 1 )} + \frac{2 t - 4}{3 ( t ^{2} - t + 1 )}

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

\int 1 d t = t

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

\int \frac{2}{3 ( t + 1 )} d t = \frac{2}{3} ln ∣ t + 1 ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{y + 4}{y ^{2} + y} d y

\frac{d}{d x} (\frac{( x ^{5} - 5 )}{x + 1})

\int_{1}^{2} (2 x^{2} - 2) d x

\int \frac{sin ( 2 x ) \cdot sinh ( 2 x )}{2} d x

\int cos (x) (7 + 4 sin^{2} (x)) d x