integral of (x^4+x+1+x^{-1}+x^{-2})

Lösung

\int (x^{4} + x + 1 + x^{- 1} + x^{- 2}) d x

\frac{x ^{5}}{5} + \frac{x ^{2}}{2} + x + ln ∣ x ∣ - \frac{1}{x} + C

Schritte zur Lösung

\int x^{4} + x + 1 + x^{- 1} + x^{- 2} 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 x^{4} d x + \int x d x + \int 1 d x + \int x^{- 1} d x + \int x^{- 2} d x

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

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int x^{- 1} d x = ln ∣ x ∣

\int x^{- 2} d x = - \frac{1}{x}

= \frac{x ^{5}}{5} + \frac{x ^{2}}{2} + x + ln ∣ x ∣ - \frac{1}{x}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{5}}{5} + \frac{x ^{2}}{2} + x + ln ∣ x ∣ - \frac{1}{x} + C

\int_{9}^{16} \frac{x}{x - 4} d x

d er i v a t i v e 7 t - 7 e^{t}

d er i v a t i v e f (x) = ∣ 4 + 7 x ∣

l o n g d i v i s i o n \frac{2 x - 1}{2 x + 1}

d er i v a t i v e f (x) = (x + 6)^{2}