1. **Problem statement:** Evaluate the indefinite integral $$\int \frac{\tan^{-1}(x)}{x} \, dx$$ as a power series and find the radius of convergence $R$. 2. **Recall the power series for arctan(x):** $$\tan^{-1}(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$$ This series converges for $$|x| < 1$$. 3. **Divide by $x$ inside the integral:** $$\frac{\tan^{-1}(x)}{x} = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \cdot \frac{1}{x} = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{2n+1}$$ 4. **Integrate term-by-term:** $$\int \frac{\tan^{-1}(x)}{x} \, dx = \int \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{2n+1} \, dx = \sum_{n=0}^\infty (-1)^n \frac{1}{2n+1} \int x^{2n} \, dx$$ 5. **Integrate each term:** $$\int x^{2n} \, dx = \frac{x^{2n+1}}{2n+1} + C$$ 6. **Substitute back:** $$\int \frac{\tan^{-1}(x)}{x} \, dx = C + \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)^2}$$ 7. **Radius of convergence:** Since the original series for $\tan^{-1}(x)$ converges for $|x|<1$, and dividing by $x$ and integrating term-by-term does not change the radius, the radius of convergence is $$R = 1$$ **Final answer:** $$f(x) = C + \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)^2}$$ $$R = 1$$