integral of (2520x)\sqrt[3]{5+3x}

Lösung

\int (2520 x) 3 5 + 3 x d x

280 (\frac{3}{7} (3 x + 5)^{\frac{7}{3}} - \frac{15}{4} (3 x + 5)^{\frac{4}{3}}) + C

Schritte zur Lösung

\int 2520 x 3 5 + 3 x d x

Entferne die Konstante:

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

= 2520 \cdot \int x 3 5 + 3 x d x

Wende U-Substitution an

= 2520 \cdot \int \frac{3 u ( u - 5 )}{9} d u

Entferne die Konstante:

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

= 2520 \cdot \frac{1}{9} \cdot \int 3 u (u - 5) d u

Multipliziere aus

3 u (u - 5) : u^{\frac{4}{3}} - 5 3 u

= 2520 \cdot \frac{1}{9} \cdot \int u^{\frac{4}{3}} - 5 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

= 2520 \cdot \frac{1}{9} (\int u^{\frac{4}{3}} d u - \int 5 3 u d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int 5 3 u d u = \frac{15}{4} u^{\frac{4}{3}}

= 2520 \cdot \frac{1}{9} (\frac{3}{7} u^{\frac{7}{3}} - \frac{15}{4} u^{\frac{4}{3}})

Setze in

u = 5 + 3 x

ein

= 2520 \cdot \frac{1}{9} (\frac{3}{7} (5 + 3 x)^{\frac{7}{3}} - \frac{15}{4} (5 + 3 x)^{\frac{4}{3}})

Vereinfache

2520 \cdot \frac{1}{9} (\frac{3}{7} (5 + 3 x)^{\frac{7}{3}} - \frac{15}{4} (5 + 3 x)^{\frac{4}{3}}) : 280 (\frac{3}{7} (3 x + 5)^{\frac{7}{3}} - \frac{15}{4} (3 x + 5)^{\frac{4}{3}})

= 280 (\frac{3}{7} (3 x + 5)^{\frac{7}{3}} - \frac{15}{4} (3 x + 5)^{\frac{4}{3}})

Füge eine Konstante zur Lösung hinzu

= 280 (\frac{3}{7} (3 x + 5)^{\frac{7}{3}} - \frac{15}{4} (3 x + 5)^{\frac{4}{3}}) + C

\frac{d ^{2}}{d x ^{2}} (\frac{1 - 4 x}{1 + 4 x})

\frac{d y}{d x} = \frac{e ^{x}}{y + 2}

\frac{\partial}{\partial x} (x^{2} - 7 x + 12)

\int \frac{x}{x ^{4} - 16} d x

x^{2} y^{^{''}} - 7 x y^{^{'}} + 12 y = 0