integral of x/(1+4x)

Lösung

\int \frac{x}{1 + 4 x} d x

\frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣) + C

Schritte zur Lösung

\int \frac{x}{1 + 4 x} d x

Wende U-Substitution an

= \int \frac{u - 1}{16 u} d u

Entferne die Konstante:

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

= \frac{1}{16} \cdot \int \frac{u - 1}{u} d u

Multipliziere aus

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

= \frac{1}{16} \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}{16} (\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}{16} (u - ln ∣ u ∣)

Setze in

u = 1 + 4 x

ein

= \frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣) + C

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