1. **State the problem:** We need to find the integral $$\int \arctan\left(\sqrt[3]{6x} - 1\right) \, dx.$$\n\n2. **Recall the formula for integration by parts:** \n$$\int u \, dv = uv - \int v \, du.$$\nWe will use integration by parts because the integrand is a product of a function and its derivative is simpler to handle.\n\n3. **Choose parts:** \nLet $u = \arctan\left(\sqrt[3]{6x} - 1\right)$ and $dv = dx$.\nThen $du = \frac{1}{1 + \left(\sqrt[3]{6x} - 1\right)^2} \cdot \frac{d}{dx}\left(\sqrt[3]{6x} - 1\right) dx$ and $v = x$.\n\n4. **Compute $du$:** \nFirst, $\frac{d}{dx}\left(\sqrt[3]{6x} - 1\right) = \frac{d}{dx}\left((6x)^{1/3} - 1\right) = \frac{1}{3}(6x)^{-2/3} \cdot 6 = \frac{2}{(6x)^{2/3}}.$\nSo, \n$$du = \frac{1}{1 + \left(\sqrt[3]{6x} - 1\right)^2} \cdot \frac{2}{(6x)^{2/3}} dx.$$\n\n5. **Apply integration by parts:** \n$$\int \arctan\left(\sqrt[3]{6x} - 1\right) dx = x \cdot \arctan\left(\sqrt[3]{6x} - 1\right) - \int x \cdot \frac{2}{(6x)^{2/3} \left[1 + \left(\sqrt[3]{6x} - 1\right)^2\right]} dx.$$\n\n6. **Simplify the integral:** \nRewrite $x / (6x)^{2/3} = x / (6^{2/3} x^{2/3}) = \frac{x^{1 - 2/3}}{6^{2/3}} = \frac{x^{1/3}}{6^{2/3}}.$\nSo the integral becomes \n$$\int \frac{2 x^{1/3}}{6^{2/3} \left[1 + \left(\sqrt[3]{6x} - 1\right)^2\right]} dx.$$\n\n7. **Substitute $t = \sqrt[3]{6x} - 1$:** \nThen $\sqrt[3]{6x} = t + 1$, so $6x = (t + 1)^3$, and $x = \frac{(t + 1)^3}{6}.$\nAlso, $dx = \frac{3(t + 1)^2}{6} dt = \frac{(t + 1)^2}{2} dt.$\n\n8. **Rewrite the integral in terms of $t$:** \nNote $x^{1/3} = \left(\frac{(t + 1)^3}{6}\right)^{1/3} = \frac{t + 1}{6^{1/3}}.$\nSo the integral is \n$$\int \frac{2 \cdot \frac{t + 1}{6^{1/3}}}{6^{2/3} (1 + t^2)} \cdot \frac{(t + 1)^2}{2} dt = \int \frac{(t + 1)^3}{6^{1/3 + 2/3} (1 + t^2)} dt = \int \frac{(t + 1)^3}{6 (1 + t^2)} dt.$$\n\n9. **Expand numerator:** \n$$(t + 1)^3 = t^3 + 3t^2 + 3t + 1.$$\nSo the integral is \n$$\frac{1}{6} \int \frac{t^3 + 3t^2 + 3t + 1}{1 + t^2} dt.$$\n\n10. **Divide the polynomial:** \nDivide numerator by denominator: \n$$\frac{t^3 + 3t^2 + 3t + 1}{1 + t^2} = t + 3 + \frac{t + 1}{1 + t^2}.$$\n\n11. **Split the integral:** \n$$\frac{1}{6} \int \left(t + 3 + \frac{t + 1}{1 + t^2}\right) dt = \frac{1}{6} \left(\int t dt + \int 3 dt + \int \frac{t + 1}{1 + t^2} dt\right).$$\n\n12. **Integrate each term:** \n- $\int t dt = \frac{t^2}{2}$\n- $\int 3 dt = 3t$\n- $\int \frac{t}{1 + t^2} dt = \frac{1}{2} \ln(1 + t^2)$ (using substitution)\n- $\int \frac{1}{1 + t^2} dt = \arctan t$\n\nSo, \n$$\int \frac{t + 1}{1 + t^2} dt = \frac{1}{2} \ln(1 + t^2) + \arctan t.$$\n\n13. **Combine all:** \n$$\frac{1}{6} \left( \frac{t^2}{2} + 3t + \frac{1}{2} \ln(1 + t^2) + \arctan t \right) + C.$$\n\n14. **Recall substitution:** \nReplace $t = \sqrt[3]{6x} - 1$ back:\n$$\int \arctan\left(\sqrt[3]{6x} - 1\right) dx = x \arctan\left(\sqrt[3]{6x} - 1\right) - \frac{1}{6} \left( \frac{(\sqrt[3]{6x} - 1)^2}{2} + 3(\sqrt[3]{6x} - 1) + \frac{1}{2} \ln\left(1 + (\sqrt[3]{6x} - 1)^2\right) + \arctan\left(\sqrt[3]{6x} - 1\right) \right) + C.$$