1. **State the problem:** We want to evaluate the integral $$\int \frac{x \, dx}{\sqrt{49 + x^4}}.$$\n\n2. **Identify the integral type and substitution:** Notice the expression under the square root is $49 + x^4$. A useful substitution is to let $t = x^2$, so that $dt = 2x \, dx$ or equivalently $x \, dx = \frac{dt}{2}$.\n\n3. **Rewrite the integral in terms of $t$:**\n$$\int \frac{x \, dx}{\sqrt{49 + x^4}} = \int \frac{1}{\sqrt{49 + t^2}} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{dt}{\sqrt{49 + t^2}}.$$\n\n4. **Recall the standard integral formula:**\n$$\int \frac{dt}{\sqrt{a^2 + t^2}} = \ln \left| t + \sqrt{a^2 + t^2} \right| + C,$$ where $a$ is a constant. Here, $a^2 = 49$, so $a = 7$.\n\n5. **Apply the formula:**\n$$\frac{1}{2} \int \frac{dt}{\sqrt{49 + t^2}} = \frac{1}{2} \ln \left| t + \sqrt{49 + t^2} \right| + C.$$\n\n6. **Substitute back $t = x^2$:**\n$$\int \frac{x \, dx}{\sqrt{49 + x^4}} = \frac{1}{2} \ln \left| x^2 + \sqrt{49 + x^4} \right| + C.$$\n\n**Final answer:**\n$$\boxed{\int \frac{x \, dx}{\sqrt{49 + x^4}} = \frac{1}{2} \ln \left| x^2 + \sqrt{49 + x^4} \right| + C}.$$