We will evaluate the integral:

$I = \int \frac{4t + 8}{\sqrt[3]{4t} + t^2} \, dt$

Step 1: Substitution

Let: $u = \sqrt[3]{4t} = (4t)^{1/3}$ Now, differentiate both sides:

$\frac{du}{dt} = \frac{d}{dt} (4t)^{1/3}$

Using the power rule:

$\frac{du}{dt} = \frac{1}{3} (4t)^{-2/3} \cdot 4 = \frac{4}{3} (4t)^{-2/3} = \frac{4}{3} u^{-2}$

Thus,

$du = \frac{4}{3} u^{-2} dt$

Now express $dt$:

$dt = \frac{3}{4} u^2 \, du$

Step 2: Rewrite the Integral

Using $u = (4t)^{1/3}$, rewrite the denominator:

$\sqrt[3]{4t} + t^2 = u + t^2$

Since $t = \frac{u^3}{4}$, we substitute:

$t^2 = \left(\frac{u^3}{4} \right)^2 = \frac{u^6}{16}$

Thus, the denominator becomes:

$u + \frac{u^6}{16}$

Similarly, rewrite the numerator:

$4t + 8 = 4 \cdot \frac{u^3}{4} + 8 = u^3 + 8$

So the integral transforms into:

$I = \int \frac{u^3 + 8}{u + \frac{u^6}{16}} \cdot \frac{3}{4} u^2 \, du$

Step 3: Simplify and Integrate

Expanding the fraction and solving the integral requires algebraic manipulation, potentially breaking into partial fractions.

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