integral of (6(x+1))/(sqrt(2x+1))

Lösung

\int \frac{6 ( x + 1 )}{2 x + 1} d x

6 (\frac{1}{3} x 2 x + 1 + \frac{2}{3} 2 x + 1) + C

Schritte zur Lösung

\int \frac{6 ( x + 1 )}{2 x + 1} d x

Entferne die Konstante:

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

= 6 \cdot \int \frac{x + 1}{2 x + 1} d x

Multipliziere aus

\frac{x + 1}{2 x + 1} : \frac{x}{2 x + 1} + \frac{1}{2 x + 1}

= 6 \cdot \int \frac{x}{2 x + 1} + \frac{1}{2 x + 1} d x

Wende die Summenregel an:

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

= 6 (\int \frac{x}{2 x + 1} d x + \int \frac{1}{2 x + 1} d x)

\int \frac{x}{2 x + 1} d x = \frac{1}{3} (x - 1) 2 x + 1

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

= 6 (\frac{1}{3} (x - 1) 2 x + 1 + 2 x + 1)

Multipliziere aus

\frac{1}{3} (x - 1) 2 x + 1 + 2 x + 1 : \frac{1}{3} x 2 x + 1 + \frac{2}{3} 2 x + 1

= 6 (\frac{1}{3} x 2 x + 1 + \frac{2}{3} 2 x + 1)

Füge eine Konstante zur Lösung hinzu

= 6 (\frac{1}{3} x 2 x + 1 + \frac{2}{3} 2 x + 1) + C

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