1. **State the problem:** We want to find the integral $$\int \frac{\tan x}{x} \, dx$$. 2. **Understand the integral:** This integral involves the function $$\frac{\tan x}{x}$$ which is not straightforward to integrate using elementary functions. 3. **Recall:** There is no elementary antiderivative for $$\frac{\tan x}{x}$$. It is typically expressed in terms of special functions or evaluated numerically. 4. **Approach:** We can express $$\tan x$$ as its Taylor series expansion: $$\tan x = \sum_{n=1}^\infty (-1)^{n-1} \frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!} x^{2n-1}$$ where $$B_{2n}$$ are Bernoulli numbers. 5. **Divide by $$x$$:** $$\frac{\tan x}{x} = \sum_{n=1}^\infty (-1)^{n-1} \frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!} x^{2n-2}$$ 6. **Integrate term-by-term:** $$\int \frac{\tan x}{x} \, dx = C + \sum_{n=1}^\infty (-1)^{n-1} \frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!} \int x^{2n-2} \, dx$$ 7. **Integrate each term:** $$\int x^{2n-2} \, dx = \frac{x^{2n-1}}{2n-1} + C$$ 8. **Final series form:** $$\int \frac{\tan x}{x} \, dx = C + \sum_{n=1}^\infty (-1)^{n-1} \frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)! (2n-1)} x^{2n-1}$$ **Summary:** The integral $$\int \frac{\tan x}{x} \, dx$$ does not have a closed form in elementary functions but can be expressed as an infinite series involving Bernoulli numbers. **Note:** For practical purposes, numerical integration or software tools are used to evaluate this integral for specific values of $$x$$.