integral of (7-6x^3+4x^6)/(x^6)

Lösung

\int \frac{7 - 6 x ^{3} + 4 x ^{6}}{x ^{6}} d x

- \frac{7}{5 x ^{5}} + \frac{3}{x ^{2}} + 4 x + C

Schritte zur Lösung

\int \frac{7 - 6 x ^{3} + 4 x ^{6}}{x ^{6}} d x

Multipliziere aus

\frac{7 - 6 x ^{3} + 4 x ^{6}}{x ^{6}} : \frac{7}{x ^{6}} - \frac{6}{x ^{3}} + 4

= \int \frac{7}{x ^{6}} - \frac{6}{x ^{3}} + 4 d x

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{7}{x ^{6}} d x - \int \frac{6}{x ^{3}} d x + \int 4 d x

\int \frac{7}{x ^{6}} d x = - \frac{7}{5 x ^{5}}

\int \frac{6}{x ^{3}} d x = - \frac{3}{x ^{2}}

\int 4 d x = 4 x

= - \frac{7}{5 x ^{5}} - (- \frac{3}{x ^{2}}) + 4 x

Vereinfache

= - \frac{7}{5 x ^{5}} + \frac{3}{x ^{2}} + 4 x

Füge eine Konstante zur Lösung hinzu

= - \frac{7}{5 x ^{5}} + \frac{3}{x ^{2}} + 4 x + C

\frac{d}{d x} (arctan (\frac{1}{3 x}))

\frac{d}{d r} (\frac{r}{r ^{2} + 5})

t \to 3 lim (\frac{2 t - 3}{t - 3})

s l o p e y = 5 sec (x) - 10 cos (x), (\frac{π}{3}, 5)

d er i v a t i v e y = arcsin (3 5 x)