What is \int(x^{x})^{2}(1+\ln x)\,dx equal to ?
- A. x^{2x}+c
- B. \frac{1}{2}x^{2x}+c ✓
- C. 2x^{2x}+c
- D. \frac{1}{2}x^{x}+c
Correct Answer: B. \frac{1}{2}x^{2x}+c
Explanation
Rewrite the integral as \int x^{2x}(1+\ln x)\,dx. Let t = x^x. Taking the natural logarithm gives \ln t = x \ln x. Differentiating both sides with respect to x gives \frac{1}{t} \frac{dt}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = 1 + \ln x. So, dt = x^x(1+\ln x)\,dx. The integral becomes \int x^x \cdot x^x (1+\ln x)\,dx = \int t \,dt = \frac{t^2}{2} + c = \frac{1}{2}x^{2x} + c.
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