1. **State the problem:** We need to find the integral $$\int \frac{x}{x^2 + 1} \, dx.$$\n\n2. **Recall the formula and rules:** When integrating a rational function where the numerator is the derivative of the denominator (or a constant multiple), substitution is a useful method. Here, notice that the derivative of the denominator $x^2 + 1$ is $2x$, which is closely related to the numerator $x$.\n\n3. **Use substitution:** Let $$u = x^2 + 1.$$ Then, $$du = 2x \, dx \implies \frac{du}{2} = x \, dx.$$\n\n4. **Rewrite the integral:** Substitute into the integral: $$\int \frac{x}{x^2 + 1} \, dx = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{u} \, du.$$\n\n5. **Integrate:** The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u| + C$. So, $$\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C.$$\n\n6. **Back-substitute:** Replace $u$ with $x^2 + 1$: $$\frac{1}{2} \ln|x^2 + 1| + C.$$\n\n7. **Simplify and conclude:** Since $x^2 + 1 > 0$ for all real $x$, we can drop the absolute value: $$\boxed{\frac{1}{2} \ln(x^2 + 1) + C}.$$\n\nThis is the final answer for the integral.