What is the integral of (2520x)\sqrt[3]{5+3x} ?

The integral of (2520x)\sqrt[3]{5+3x} is 280(3/7 (3x+5)^{7/3}-15/4 (3x+5)^{4/3})+C

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

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

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

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

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

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

Substitute back

u = 5 + 3 x

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

Simplify

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}})

Add a constant to the solution

= 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

What is the integral of (2520x)\sqrt[3]{5+3x} ?
The integral of (2520x)\sqrt[3]{5+3x} is 280(3/7 (3x+5)^{7/3}-15/4 (3x+5)^{4/3})+C