1. **State the problem:** We want to evaluate the integral $$\int_0^{1} \frac{x^x (x^{2x} + 1) (\ln x + 1)}{x^{4x} + 1} \, dx.$$\n\n2. **Analyze the integrand:** The integrand is $$\frac{x^x (x^{2x} + 1) (\ln x + 1)}{x^{4x} + 1}.$$ Notice the powers of $x$ in the numerator and denominator and the presence of $\ln x + 1$.\n\n3. **Use substitution:** Let $$y = x^x.$$ Then, taking the natural logarithm, $$\ln y = x \ln x.$$ Differentiating both sides with respect to $x$:\n$$\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1).$$\n\n4. **Rewrite the integral:** The numerator contains $x^x (\ln x + 1)$ which is exactly $\frac{dy}{dx}$. Also, note that $$x^{2x} = (x^x)^2 = y^2,$$ and $$x^{4x} = (x^x)^4 = y^4.$$\n\nSo the integrand becomes:\n$$\frac{y (y^2 + 1) (\ln x + 1)}{y^4 + 1} = \frac{(y^2 + 1)}{y^4 + 1} \cdot y (\ln x + 1) = \frac{y^2 + 1}{y^4 + 1} \cdot \frac{dy}{dx}.$$\n\n5. **Change variable in integral:** Since $dy = \frac{dy}{dx} dx$, the integral becomes:\n$$\int_{x=0}^{x=1} \frac{y^2 + 1}{y^4 + 1} dy.$$\n\n6. **Find new limits:** When $x \to 0^+$, $y = x^x = e^{x \ln x}$. Since $x \ln x \to 0$, $y \to 1$. When $x=1$, $y=1^1=1$. So the integral limits in $y$ are from 1 to 1.\n\n7. **Evaluate the integral:** Since the limits are the same, the integral evaluates to 0.\n\n**Final answer:** $$\boxed{0}.$$