1. We are asked to find the exact value of the integral $$\int_0^1 x \tan^{-1}(x) \, dx.$$\n\n2. To solve this, we use integration by parts. Recall the formula: $$\int u \, dv = uv - \int v \, du.$$\n\n3. Let $$u = \tan^{-1}(x)$$ and $$dv = x \, dx.$$ Then, $$du = \frac{1}{1+x^2} \, dx$$ and $$v = \frac{x^2}{2}.$$\n\n4. Applying integration by parts: $$\int_0^1 x \tan^{-1}(x) \, dx = \left. \frac{x^2}{2} \tan^{-1}(x) \right|_0^1 - \int_0^1 \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx.$$\n\n5. Evaluate the boundary term: $$\left. \frac{x^2}{2} \tan^{-1}(x) \right|_0^1 = \frac{1}{2} \cdot \frac{\pi}{4} - 0 = \frac{\pi}{8}.$$\n\n6. Simplify the integral: $$\int_0^1 \frac{x^2}{2(1+x^2)} \, dx = \frac{1}{2} \int_0^1 \frac{x^2}{1+x^2} \, dx.$$\n\n7. Rewrite the integrand: $$\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}.$$\n\n8. So, $$\frac{1}{2} \int_0^1 \left(1 - \frac{1}{1+x^2}\right) dx = \frac{1}{2} \left( \int_0^1 1 \, dx - \int_0^1 \frac{1}{1+x^2} \, dx \right).$$\n\n9. Evaluate each integral: $$\int_0^1 1 \, dx = 1,$$ and $$\int_0^1 \frac{1}{1+x^2} \, dx = \left. \tan^{-1}(x) \right|_0^1 = \frac{\pi}{4}.$$\n\n10. Substitute back: $$\frac{1}{2} (1 - \frac{\pi}{4}) = \frac{1}{2} - \frac{\pi}{8}.$$\n\n11. Finally, combine all parts: $$\int_0^1 x \tan^{-1}(x) \, dx = \frac{\pi}{8} - \left( \frac{1}{2} - \frac{\pi}{8} \right) = \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2}.$$\n\nAnswer: $$\boxed{\frac{\pi}{4} - \frac{1}{2}}.$$