To compute the integral of $\sqrt[3]{x^2}$ (cube root of $x^2$), we can write it in exponential form:

$\cbrt{x^2} = x^{2/3}.$

The integral becomes:

$\int x^{2/3} \, dx.$

Using the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (valid for $n \neq -1$):

$\int x^{2/3} \, dx = \frac{x^{2/3 + 1}}{2/3 + 1} + C.$

Simplify the exponent $2/3 + 1 = 5/3$:

$\int x^{2/3} \, dx = \frac{x^{5/3}}{5/3} + C = \frac{3}{5}x^{5/3} + C.$

Final Answer:

$\int \cbrt{x^2} \, dx = \frac{3}{5}x^{5/3} + C.$

Would you like to explore related integrals or have this broken down further?

Related Questions:

1. What is the integral of $\sqrt{x^2}$?
2. How is the power rule for integration derived?
3. Can you solve $\int x^{a} \, dx$ where $a = -2/3$?
4. What happens when $n = -1$ in the power rule for integration?
5. How would the integral change if $\cbrt{x^2}$ were in a denominator?

Tip:

Always rewrite roots and radicals into fractional exponents for easier integration or differentiation.