To solve the integral
$\int (\sqrt[3]{x})^4 \, dx$
we will proceed as follows:

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Step 1: Simplify the exponent

The term $(\sqrt[3]{x})^4$ can be rewritten using properties of exponents. Recall that:

$\sqrt[3]{x} = x^{1/3}.$

Raising this to the 4th power:

$(\sqrt[3]{x})^4 = (x^{1/3})^4 = x^{4/3}.$

Thus, the integral becomes:

$\int x^{4/3} \, dx.$

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Step 2: Apply the power rule for integration

The power rule for integration states:

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad \text{where n \neq -1}.

Here, $n = \frac{4}{3}$. Adding 1 to the exponent:

$n + 1 = \frac{4}{3} + 1 = \frac{4}{3} + \frac{3}{3} = \frac{7}{3}.$

Now, apply the power rule:

$\int x^{4/3} \, dx = \frac{x^{7/3}}{7/3} + C.$

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Step 3: Simplify the result

Dividing by $\frac{7}{3}$ is equivalent to multiplying by its reciprocal:

$\frac{x^{7/3}}{7/3} = \frac{3}{7} x^{7/3}.$

Thus, the solution is:

$\int (\sqrt[3]{x})^4 \, dx = \frac{3}{7} x^{7/3} + C.$

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Final Answer:

$\boxed{\frac{3}{7} x^{7/3} + C.}$

Let me know if you need further clarification!

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Related Questions:

1. How do you integrate functions with fractional exponents?
2. What is the general formula for integrating $x^n$ when $n \neq -1$?
3. How do you differentiate $x^{4/3}$ to verify the result?
4. How would the integral change if the exponent were $(\sqrt[3]{x})^3$?
5. What are the steps to solve definite integrals involving fractional powers?

Tip:

When dealing with radicals, always rewrite them as fractional exponents to simplify integration or differentiation.