integral of (2x)/(5x+1)

Lösung

\int \frac{2 x}{5 x + 1} d x

\frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣) + C

Schritte zur Lösung

\int \frac{2 x}{5 x + 1} d x

Entferne die Konstante:

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

= 2 \cdot \int \frac{x}{5 x + 1} d x

Wende U-Substitution an

= 2 \cdot \int \frac{u - 1}{25 u} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{25} \cdot \int \frac{u - 1}{u} d u

Multipliziere aus

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

= 2 \cdot \frac{1}{25} \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

= 2 \cdot \frac{1}{25} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{25} (u - ln ∣ u ∣)

Setze in

u = 5 x + 1

ein

= 2 \cdot \frac{1}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣)

Vereinfache

2 \cdot \frac{1}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣) : \frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣)

= \frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣) + C

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