integral of x/(4x-3)

Lösung

\int \frac{x}{4 x - 3} d x

\frac{1}{16} (4 x - 3 + 3 ln ∣ 4 x - 3 ∣) + C

Schritte zur Lösung

\int \frac{x}{4 x - 3} d x

Wende U-Substitution an

= \int \frac{u + 3}{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 + 3}{u} d u

Multipliziere aus

\frac{u + 3}{u} : 1 + \frac{3}{u}

= \frac{1}{16} \cdot \int 1 + \frac{3}{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{3}{u} d u)

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= \frac{1}{16} (u + 3 ln ∣ u ∣)

Setze in

u = 4 x - 3

ein

= \frac{1}{16} (4 x - 3 + 3 ln ∣ 4 x - 3 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (4 x - 3 + 3 ln ∣ 4 x - 3 ∣) + C

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