integral of x^{-8}(x+1)

Lösung

\int x^{- 8} (x + 1) d x

- \frac{1}{6 x ^{6}} - \frac{1}{7 x ^{7}} + C

Schritte zur Lösung

\int x^{- 8} (x + 1) d x

Vereinfache

x^{- 8} (x + 1) : \frac{x + 1}{x ^{8}}

= \int \frac{x + 1}{x ^{8}} d x

Multipliziere aus

\frac{x + 1}{x ^{8}} : \frac{1}{x ^{7}} + \frac{1}{x ^{8}}

= \int \frac{1}{x ^{7}} + \frac{1}{x ^{8}} 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{1}{x ^{7}} d x + \int \frac{1}{x ^{8}} d x

\int \frac{1}{x ^{7}} d x = - \frac{1}{6 x ^{6}}

\int \frac{1}{x ^{8}} d x = - \frac{1}{7 x ^{7}}

= - \frac{1}{6 x ^{6}} - \frac{1}{7 x ^{7}}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6 x ^{6}} - \frac{1}{7 x ^{7}} + C

d er i v a t i v e \frac{sin ( 8 x )}{13 x}

d er i v a t i v e 48 x (4 x^{2} + 9)^{5}

d er i v a t i v e (t^{2} - 4 t) (t^{2} + 4 t)

(x + 1)^{2} y^{^{'}} + 5 (x + 1) y = 16

i = 1 \sum \infty \frac{1}{i ^{2}}