1. **State the problem:** Evaluate the integral $$\int_1^{\sqrt{3}} \tan^{-1}\left(\frac{1}{x}\right) \, dx.$$\n\n2. **Substitution:** Let $$u = \frac{1}{x}$$ so that $$x = \frac{1}{u}$$ and $$dx = -\frac{1}{u^2} du.$$\n\n3. **Change limits:** When $$x=1$$, $$u=1$$; when $$x=\sqrt{3}$$, $$u=\frac{1}{\sqrt{3}}.$$\n\n4. **Rewrite the integral:**\n$$\int_1^{\sqrt{3}} \tan^{-1}\left(\frac{1}{x}\right) dx = \int_1^{\sqrt{3}} \tan^{-1}(u) dx = \int_1^{\frac{1}{\sqrt{3}}} \tan^{-1}(u) \left(-\frac{1}{u^2}\right) du = -\int_1^{\frac{1}{\sqrt{3}}} \frac{\tan^{-1}(u)}{u^2} du.$$\n\n5. **Reverse limits to remove negative sign:**\n$$-\int_1^{\frac{1}{\sqrt{3}}} \frac{\tan^{-1}(u)}{u^2} du = \int_{\frac{1}{\sqrt{3}}}^1 \frac{\tan^{-1}(u)}{u^2} du.$$\n\n6. **Integration by parts:** Let $$w = \tan^{-1}(u)$$ and $$dv = \frac{1}{u^2} du = u^{-2} du.$$\nThen, $$dw = \frac{1}{1+u^2} du$$ and $$v = \int u^{-2} du = -\frac{1}{u}.$$\n\n7. **Apply integration by parts formula:**\n$$\int w \, dv = wv - \int v \, dw,$$\nso\n$$\int \frac{\tan^{-1}(u)}{u^2} du = -\frac{\tan^{-1}(u)}{u} - \int \left(-\frac{1}{u}\right) \frac{1}{1+u^2} du = -\frac{\tan^{-1}(u)}{u} + \int \frac{1}{u(1+u^2)} du.$$\n\n8. **Simplify the remaining integral:**\n$$\int \frac{1}{u(1+u^2)} du = \int \left(\frac{1}{u} - \frac{u}{1+u^2}\right) du$$ because\n$$\frac{1}{u(1+u^2)} = \frac{1}{u} - \frac{u}{1+u^2}.$$\n\n9. **Integrate each term:**\n$$\int \frac{1}{u} du = \ln|u|,$$\n$$\int \frac{u}{1+u^2} du = \frac{1}{2} \ln(1+u^2)$$ (using substitution).\n\n10. **Combine results:**\n$$\int \frac{1}{u(1+u^2)} du = \ln|u| - \frac{1}{2} \ln(1+u^2) + C.$$\n\n11. **Full integral expression:**\n$$\int \frac{\tan^{-1}(u)}{u^2} du = -\frac{\tan^{-1}(u)}{u} + \ln|u| - \frac{1}{2} \ln(1+u^2) + C.$$\n\n12. **Evaluate definite integral from $$u=\frac{1}{\sqrt{3}}$$ to $$u=1$$:**\n$$\left[-\frac{\tan^{-1}(u)}{u} + \ln|u| - \frac{1}{2} \ln(1+u^2)\right]_{\frac{1}{\sqrt{3}}}^1.$$\n\n13. **Calculate values:**\n- $$\tan^{-1}(1) = \frac{\pi}{4}, \quad \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.$$\n- $$\ln(1) = 0, \quad \ln\left(\frac{1}{\sqrt{3}}\right) = -\frac{1}{2} \ln 3.$$\n- $$\ln(1+1^2) = \ln 2, \quad \ln\left(1 + \frac{1}{3}\right) = \ln \frac{4}{3}.$$\n\n14. **Substitute and simplify:**\n$$\left[-\frac{\pi/4}{1} + 0 - \frac{1}{2} \ln 2\right] - \left[-\frac{\pi/6}{\frac{1}{\sqrt{3}}} - \frac{1}{2} \ln 3 - \frac{1}{2} \ln \frac{4}{3}\right].$$\n\n15. **Simplify terms:**\n$$-\frac{\pi}{4} - \frac{1}{2} \ln 2 - \left[-\pi \frac{\sqrt{3}}{6} - \frac{1}{2} \ln 3 - \frac{1}{2} \ln \frac{4}{3}\right].$$\n\n16. **Note:** $$\frac{\pi}{6} \times \sqrt{3} = \frac{\pi \sqrt{3}}{6}$$ so\n$$-\frac{\pi}{4} - \frac{1}{2} \ln 2 + \frac{\pi \sqrt{3}}{6} + \frac{1}{2} \ln 3 + \frac{1}{2} \ln \frac{4}{3}.$$\n\n17. **Combine logarithms:**\n$$\frac{1}{2} \ln 3 + \frac{1}{2} \ln \frac{4}{3} - \frac{1}{2} \ln 2 = \frac{1}{2} \ln \left(3 \times \frac{4}{3} \times \frac{1}{2}\right) = \frac{1}{2} \ln 2.$$\n\n18. **Final answer:**\n$$\int_1^{\sqrt{3}} \tan^{-1}\left(\frac{1}{x}\right) dx = \frac{\pi \sqrt{3}}{6} - \frac{\pi}{4} + \frac{1}{2} \ln 2.$$