<p>The value of the definite integral \(\displaystyle\int_{1}^{\sqrt{3}}\left(x^{2x^2+1}+\ln\left(x^{x^{\left(2x^2+1\right)}}\right)\right)dx\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that the logarithmic term simplifies to x(2x² + 1)ln(x), making the integrand a perfect derivative of x^(2x²+1) when combined with its exponential derivative form.
<p><strong>Step 1:</strong> Simplify the logarithmic term.</p><p>ln(x^(x(2x²+1))) = x(2x² + 1)ln(x)</p><p><strong>Step 2:</strong> Rewrite the integrand as:</p><p>∫[x^(2x²+1) + x(2x² + 1)ln(x)]dx</p><p><strong>Step 3:</strong> Find d/dx[x^(2x²+1)]. Using logarithmic differentiation:</p><p>Let y = x^(2x²+1), then ln(y) = (2x² + 1)ln(x)</p><p>d/dx[ln(y)] = 4x·ln(x) + (2x² + 1)·(1/x)</p><p>dy/dx = x^(2x²+1)[4x·ln(x) + (2x² + 1)/x]</p><p>dy/dx = x^(2x²+1)·4x·ln(x) + x^(2x²+1)·(2x² + 1)/x</p><p>dy/dx = 4x²·x^(2x²+1)·ln(x)/x + x^(2x²+1)·(2x² + 1)/x</p><p><strong>Step 4:</strong> Recognize the integrand as d/dx[x^(2x²+1)]</p><p>∫₁^√3 [x^(2x²+1) + x(2x² + 1)ln(x)]dx = [x^(2x²+1)]₁^√3</p><p><strong>Step 5:</strong> Evaluate at bounds:</p><p>At x = √3: (√3)^(2·3+1) = (√3)^7 = 3^(7/2) = 27√3</p><p>At x = 1: 1^(2·1+1) = 1^3 = 1</p><p>∴ Answer: 27√3 - 1 (Option C)</p>
Correct Answer: C