integral of-x/(1+x)

Lösung

\int - \frac{x}{1 + x} d x

- 1 - x + ln ∣ 1 + x ∣ + C

Schritte zur Lösung

\int - \frac{x}{1 + x} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{x}{1 + x} d x

Wende U-Substitution an

= - \int \frac{u - 1}{u} d u

Multipliziere aus

\frac{u - 1}{u} : 1 - \frac{1}{u}

= - \int 1 - \frac{1}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

Setze in

u = 1 + x

ein

= - (1 + x - ln ∣ 1 + x ∣)

Vereinfache

= - 1 - x + ln ∣ 1 + x ∣

Füge eine Konstante zur Lösung hinzu

= - 1 - x + ln ∣ 1 + x ∣ + C

a re a 6 - y = x^{2}, y + 8 = x^{2}, x - y + 4 = 0, x - y - 6 = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - 2, y^{^{'}} (0) = 4

\frac{d}{d x} (\frac{1}{x + sin ( x )})

\frac{d y}{d x} = \frac{4 e ^{4 x}}{e ^{y}}

\int 2 e^{- s t} d t