1. The problem is to evaluate the definite integral $$\int_0^1 \tan^{-1}(x)\,dx$$. 2. We use integration by parts, where we let: - $u = \tan^{-1}(x)$, so $du = \frac{1}{1+x^2} dx$ - $dv = dx$, so $v = x$ 3. Integration by parts formula is: $$\int u\,dv = uv - \int v\,du$$ 4. Applying this, we get: $$\int_0^1 \tan^{-1}(x) dx = \left. x \tan^{-1}(x) \right|_0^1 - \int_0^1 \frac{x}{1+x^2} dx$$ 5. Evaluate the first term: $$1 \cdot \tan^{-1}(1) - 0 \cdot \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$ 6. For the second integral, use substitution: Let $w = 1 + x^2$, so $dw = 2x dx$, thus $x dx = \frac{dw}{2}$ 7. Change limits for $w$: When $x=0$, $w=1$; when $x=1$, $w=2$ 8. Substitute into the integral: $$\int_0^1 \frac{x}{1+x^2} dx = \int_1^2 \frac{1}{w} \cdot \frac{dw}{2} = \frac{1}{2} \int_1^2 \frac{1}{w} dw = \frac{1}{2} \ln(w) \Big|_1^2 = \frac{1}{2} \ln 2$$ 9. Combine results: $$\int_0^1 \tan^{-1}(x) dx = \frac{\pi}{4} - \frac{1}{2} \ln 2$$ Final answer: $$\boxed{\int_0^1 \tan^{-1}(x) dx = \frac{\pi}{4} - \frac{1}{2} \ln 2}$$