Question:

Step 1: Simplify the logarithmic term in the integrand. We need to simplify $\ln\left(x^{2x^{2x^2+1}}\right)$. Using the logarithm property $\ln(a^b) = b\ln(a)$: $$\ln\left(x^{2x^{2x^2+1}}\right) = 2x^{2x^2+1} \cdot \ln(x)$$ Step 2: Rewrite the integral with the simplified expression. The integral becomes: $$I = \int_{1}^{\sqrt{3}} \left[x^{2x^2+1} + 2x^{2x^2+1} \cdot \ln(x)\right] dx$$ Step 3: Factor out the common term. We can factor $x^{2x^2+1}$ from both terms: $$I = \int_{1}^{\sqrt{3}} x^{2x^2+1}(1 + 2\ln x) \, dx$$ Step 4: Find the derivative of $x^{2x^2}$ using logarithmic differentiation. Let $y = x^{2x^2}$. Taking the natural logarithm of both sides: $$\ln y = 2x^2 \ln x$$ Differentiating both sides with respect to $x$: $$\frac{1}{y}\frac{dy}{dx} = 2x^2 \cdot \frac{1}{x} + \ln x \cdot 4x = 2x + 4x\ln x = 2x(1 + 2\ln x)$$ Therefore: $$\frac{dy}{dx} = x^{2x^2} \cdot 2x(1 + 2\ln x) = 2x^{2x^2+1}(1 + 2\ln x)$$ Step 5: Express the integrand as a derivative. From Step 4, we have: $$x^{2x^2+1}(1 + 2\ln x) = \frac{1}{2} \cdot \frac{d}{dx}\left[x^{2x^2}\right]$$ Step 6: Evaluate the definite integral using the Fundamental Theorem of Calculus. $$I = \int_{1}^{\sqrt{3}} x^{2x^2+1}(1 + 2\ln x) \, dx = \frac{1}{2}\left[x^{2x^2}\right]_{1}^{\sqrt{3}}$$ $$= \frac{1}{2}\left[(\sqrt{3})^{2(\sqrt{3})^2} - 1^{2 \cdot 1^2}\right]$$ $$= \frac{1}{2}\left[(\sqrt{3})^{2 \cdot 3} - 1\right]$$ $$= \frac{1}{2}\left[(\sqrt{3})^6 - 1\right]$$ Step 7: Simplify the expression. Note that $(\sqrt{3})^6 = (3^{1/2})^6 = 3^3 = 27$: $$I = \frac{1}{2}(27 - 1) = \frac{1}{2}(26) = 13$$ **Final Answer:** The value of the definite integral is $\boxed{13}$, which corresponds to **Option 4**.