integral of x^{-5}(x+1)

Lösung

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

- \frac{1}{3 x ^{3}} - \frac{1}{4 x ^{4}} + C

Schritte zur Lösung

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

Vereinfache

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

= \int \frac{x + 1}{x ^{5}} d x

Multipliziere aus

\frac{x + 1}{x ^{5}} : \frac{1}{x ^{4}} + \frac{1}{x ^{5}}

= \int \frac{1}{x ^{4}} + \frac{1}{x ^{5}} 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 ^{4}} d x + \int \frac{1}{x ^{5}} d x

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

= - \frac{1}{3 x ^{3}} - \frac{1}{4 x ^{4}}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{3 x ^{3}} - \frac{1}{4 x ^{4}} + C

\frac{\partial}{\partial y} (x^{2} + y^{2} - x y)

d er i v a t i v e x e^{x - 1}

\int e^{7 x} d x

\frac{d}{d x} (x e)

f (x) = 13 x