integral of x/(2x-1)

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u + 1}{4 u} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int \frac{u + 1}{u} d u

Multipliziere aus

\frac{u + 1}{u} : 1 + \frac{1}{u}

= \frac{1}{4} \cdot \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

= \frac{1}{4} (\int 1 d u + \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{4} (u + ln ∣ u ∣)

Setze in

u = 2 x - 1

ein

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

Füge eine Konstante zur Lösung hinzu

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

y^{^{'}} = \frac{x}{10 y}

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

\frac{\partial}{\partial x} (x + x y + y)

y^{^{'}} = (1 - 2 x) y^{2}

\frac{d}{d x} (4 x e^{2 x})