1. **State the problem:** Evaluate the integral $$\int \frac{\tan^{-1} \left( \frac{1}{x} \right)}{1 + x^2} \, dx$$. 2. **Recall the formula and substitution:** Let $$y = \tan^{-1} \left( \frac{1}{x} \right)$$. We want to integrate $$\frac{y}{1+x^2}$$ with respect to $$x$$. 3. **Use substitution and differentiation:** Note that $$\frac{d}{dx} \left( \tan^{-1} x \right) = \frac{1}{1+x^2}$$. Also, $$\tan^{-1} \left( \frac{1}{x} \right)$$ is a function of $$x$$. 4. **Integration by parts approach:** Set: $$u = \tan^{-1} \left( \frac{1}{x} \right), \quad dv = \frac{1}{1+x^2} dx$$ Then: $$du = \frac{d}{dx} \tan^{-1} \left( \frac{1}{x} \right) dx = \frac{-1/x^2}{1 + (1/x)^2} dx = \frac{-1/x^2}{1 + 1/x^2} dx = \frac{-1/x^2}{\frac{x^2+1}{x^2}} dx = \frac{-1}{x^2} \cdot \frac{x^2}{x^2+1} dx = \frac{-1}{x^2+1} dx$$ and $$v = \tan^{-1} x$$ 5. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\int \frac{\tan^{-1} \left( \frac{1}{x} \right)}{1+x^2} dx = \tan^{-1} \left( \frac{1}{x} \right) \tan^{-1} x - \int \tan^{-1} x \left( \frac{-1}{x^2+1} \right) dx$$ $$= \tan^{-1} \left( \frac{1}{x} \right) \tan^{-1} x + \int \frac{\tan^{-1} x}{1+x^2} dx$$ 6. **Recognize the integral on the right is the same as the original integral:** Let $$I = \int \frac{\tan^{-1} \left( \frac{1}{x} \right)}{1+x^2} dx$$ From above, $$I = \tan^{-1} \left( \frac{1}{x} \right) \tan^{-1} x + I$$ Subtract $$I$$ from both sides: $$I - I = \tan^{-1} \left( \frac{1}{x} \right) \tan^{-1} x$$ $$0 = \tan^{-1} \left( \frac{1}{x} \right) \tan^{-1} x$$ This suggests a contradiction, so let's try a different approach. 7. **Alternative substitution:** Rewrite the integral as: $$\int \frac{\tan^{-1} \left( \frac{1}{x} \right)}{1+x^2} dx$$ Let $$t = \tan^{-1} \left( \frac{1}{x} \right)$$. Then, $$\tan t = \frac{1}{x} \implies x = \frac{1}{\tan t} = \cot t$$ Differentiating, $$dx = -\csc^2 t \, dt$$ Also, $$1 + x^2 = 1 + \cot^2 t = \csc^2 t$$ Substitute into the integral: $$\int \frac{t}{1 + x^2} dx = \int \frac{t}{\csc^2 t} (-\csc^2 t) dt = \int -t \, dt = - \int t \, dt$$ 8. **Integrate:** $$- \int t \, dt = - \frac{t^2}{2} + C = - \frac{1}{2} \left( \tan^{-1} \left( \frac{1}{x} \right) \right)^2 + C$$ **Final answer:** $$\int \frac{\tan^{-1} \left( \frac{1}{x} \right)}{1+x^2} dx = - \frac{1}{2} \left( \tan^{-1} \left( \frac{1}{x} \right) \right)^2 + C$$