integral of-1/2 ln|x|-8x+2x^2+C^1

Lösung

\int - \frac{1}{2} ln ∣ x ∣ - 8 x + 2 x^{2} + C^{1} d x

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

Schritte zur Lösung

\int - \frac{1}{2} ln ∣ x ∣ - 8 x + 2 x^{2} + C^{1} 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}{2} ln ∣ x ∣ d x - \int 8 x d x + \int 2 x^{2} d x + \int C^{1} d x

\int \frac{1}{2} ln ∣ x ∣ d x = \frac{1}{2} (x ln ∣ x ∣ - x)

\int 8 x d x = 4 x^{2}

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

\int C^{1} d x = C x

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

Füge eine Konstante zur Lösung hinzu

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

\int tan^{5} (x) sec^{2} (x) d x

x \to - 3 lim (- - x + 4)

\frac{d y}{d x} = y ln (x)

\frac{d}{d x} (\frac{x}{x ^{2} + 6})

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