What is the integral of (5x+3)\sqrt[3]{5x^2+6x} ?

The integral of (5x+3)\sqrt[3]{5x^2+6x} is 3/(40\sqrt[3]{5)}(25x^2+30x)^{4/3}+C

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integral of (5x+3)\sqrt[3]{5x^2+6x}

Solution steps

Apply u-substitution

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

Apply the Power Rule

: \frac{3}{4} v^{\frac{4}{3}}

Substitute back

Simplify

Add a constant to the solution

x + 2, \frac{1}{1 + x}, x = 2

y^{^{''}} + 7 y^{^{'}} + 10 y = 4 t e^{4 t}

f (x) = 10 x^{2}, at x = - 2

f (x) = \frac{2 x + 1}{2 x ^{2} + 1}

\frac{10}{x ^{3}}

What is the integral of (5x+3)\sqrt[3]{5x^2+6x} ?
The integral of (5x+3)\sqrt[3]{5x^2+6x} is 3/(40\sqrt[3]{5)}(25x^2+30x)^{4/3}+C