integral of x/(sqrt(3x+4))

Lösung

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

\frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{2 ( u ^{2} - 4 )}{9} d u

Entferne die Konstante:

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

= \frac{2}{9} \cdot \int u^{2} - 4 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{2}{9} (\int u^{2} d u - \int 4 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 4 d u = 4 u

= \frac{2}{9} (\frac{u ^{3}}{3} - 4 u)

Setze in

u = 3 x + 4

ein

= \frac{2}{9} (\frac{( 3 x + 4 ) ^{3}}{3} - 4 3 x + 4)

Vereinfache

\frac{2}{9} (\frac{( 3 x + 4 ) ^{3}}{3} - 4 3 x + 4) : \frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4)

= \frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4)

Füge eine Konstante zur Lösung hinzu

= \frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4) + C

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