1. **State the problem:** Evaluate the improper integral $$\int_{-1}^1 \frac{1}{(1+x)^{2/3}} \, dx.$$\n\n2. **Identify the issue:** The integrand has a potential problem at $x = -1$ because the denominator becomes zero, making the integrand undefined there. This means the integral is improper and we must check for convergence.\n\n3. **Rewrite the integral with a limit:** To handle the improper point at $x = -1$, write the integral as a limit:\n$$\int_{-1}^1 \frac{1}{(1+x)^{2/3}} \, dx = \lim_{t \to -1^+} \int_t^1 \frac{1}{(1+x)^{2/3}} \, dx.$$\n\n4. **Find the antiderivative:** Use substitution $u = 1+x$, so $du = dx$. The integral becomes:\n$$\int \frac{1}{u^{2/3}} \, du = \int u^{-2/3} \, du.$$\n\n5. **Integrate:**\n$$\int u^{-2/3} \, du = \frac{u^{1/3}}{(1/3)} + C = 3u^{1/3} + C.$$\n\n6. **Apply limits:** Substitute back $u = 1+x$ and evaluate from $t$ to $1$:\n$$\int_t^1 \frac{1}{(1+x)^{2/3}} \, dx = 3[(1+1)^{1/3} - (1+t)^{1/3}] = 3[2^{1/3} - (1+t)^{1/3}].$$\n\n7. **Take the limit as $t \to -1^+$:**\nAs $t \to -1^+$, $(1+t)^{1/3} \to 0^{+} = 0$. So the integral becomes:\n$$\lim_{t \to -1^+} 3[2^{1/3} - (1+t)^{1/3}] = 3 \cdot 2^{1/3} - 0 = 3 \cdot 2^{1/3}.$$\n\n8. **Conclusion:** The integral converges and its value is $$3 \cdot \sqrt[3]{2}.$$\n\n**Final answer:** (a) $3 \cdot \sqrt[3]{2}$