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

The integral of (2+t^2)\sqrt[3]{6t+t^3} is 1/4 (6t+t^3)^{4/3}+C

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

\frac{1}{4} (6 t + t^{3})^{\frac{4}{3}} + C

Solution steps

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} u^{\frac{4}{3}}

= \frac{1}{3} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 6 t + t^{3}

= \frac{1}{3} \cdot \frac{3}{4} (6 t + t^{3})^{\frac{4}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{4} (6 t + t^{3})^{\frac{4}{3}} : \frac{1}{4} (6 t + t^{3})^{\frac{4}{3}}

= \frac{1}{4} (6 t + t^{3})^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{4} (6 t + t^{3})^{\frac{4}{3}} + C

\frac{d}{d y} (4 9 - y)

y^{^{'}} = 3 x (x - 1)

x \to 0 - lim (\frac{1}{x} - cot (x))

- x^{2} + 4 x

\int_{0}^{\frac{π}{3}} 4 sec^{3} (x) d x

What is the integral of (2+t^2)\sqrt[3]{6t+t^3} ?
The integral of (2+t^2)\sqrt[3]{6t+t^3} is 1/4 (6t+t^3)^{4/3}+C