integral of 9xsqrt(6x-3)

Lösung

\int 9 x 6 x - 3 d x

\frac{3 3}{5} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + C

Schritte zur Lösung

\int 9 x 6 x - 3 d x

Entferne die Konstante:

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

= 9 \cdot \int x 6 x - 3 d x

Wende U-Substitution an

= 9 \cdot \int \frac{u ( u + 3 )}{36} d u

Entferne die Konstante:

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

= 9 \cdot \frac{1}{36} \cdot \int u (u + 3) d u

Multipliziere aus

u (u + 3) : u^{\frac{3}{2}} + 3 u

= 9 \cdot \frac{1}{36} \cdot \int u^{\frac{3}{2}} + 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

= 9 \cdot \frac{1}{36} (\int u^{\frac{3}{2}} d u + \int 3 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= 9 \cdot \frac{1}{36} (\frac{2}{5} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}})

Setze in

u = 6 x - 3

ein

= 9 \cdot \frac{1}{36} (\frac{2}{5} (6 x - 3)^{\frac{5}{2}} + 2 (6 x - 3)^{\frac{3}{2}})

Vereinfache

9 \cdot \frac{1}{36} (\frac{2}{5} (6 x - 3)^{\frac{5}{2}} + 2 (6 x - 3)^{\frac{3}{2}}) : \frac{3 3}{5} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

= \frac{3 3}{5} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{3 3}{5} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + C

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