1. **State the problem:** We want to find the integral $$I = \int x \tan^{-1}(x) \, dx.$$\n\n2. **Formula and method:** To solve this integral, we use integration by parts. Recall the formula:\n$$\int u \, dv = uv - \int v \, du.$$\nWe choose:\n- $u = \tan^{-1}(x)$, so $du = \frac{1}{1+x^2} dx$.\n- $dv = x \, dx$, so $v = \frac{x^2}{2}$.\n\n3. **Apply integration by parts:**\n$$I = \frac{x^2}{2} \tan^{-1}(x) - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int \frac{x^2}{1+x^2} dx.$$\n\n4. **Simplify the integral:**\nNote that $$\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}.$$\nSo,\n$$\int \frac{x^2}{1+x^2} dx = \int 1 \, dx - \int \frac{1}{1+x^2} dx = x - \tan^{-1}(x) + C.$$\n\n5. **Substitute back:**\n$$I = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left(x - \tan^{-1}(x)\right) + C = \frac{x^2}{2} \tan^{-1}(x) - \frac{x}{2} + \frac{1}{2} \tan^{-1}(x) + C.$$\n\n6. **Final answer:**\n$$\boxed{I = \frac{x^2 + 1}{2} \tan^{-1}(x) - \frac{x}{2} + C}.$$