<p>Let \(F(a)=\displaystyle\int_0^1\frac{x^a-1}{\ln x}\,dx\) for \(a>-1\). Which are correct?</p>
Step-by-Step Solution
Key Concept: Differentiate under the integral: \partial/\partiala(xᵃ/ln x) = xᵃ. So F'(a) = \int_0^1 xᵃ dx = 1/(a+1). Integrate: F(a) = ln(a+1)+C. F(0)=0 gives C=0.
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<p><strong>Feynman's trick:</strong> $F'(a) = \displaystyle\int_0^1\frac{\partial}{\partial a}\frac{x^a-1}{\ln x}\,dx = \int_0^1 x^a\,dx = \frac{1}{a+1}$. ✓ (B)</p>
<p><strong>Integrate:</strong> $F(a)=\int\frac{da}{a+1}=\ln(a+1)+C$.</p>
<p><strong>Initial condition:</strong> $F(0)=\int_0^1\frac{x^0-1}{\ln x}dx=\int_0^1\frac{0}{\ln x}dx=0\Rightarrow C=0$. ✓ (D)</p>
<p>$\therefore F(a)=\ln(a+1)$. ✓ (C)</p>
<p><strong>A:</strong> $F'(a)=1/a$ is wrong — it's $1/(a+1)$. ✗</p>
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Correct Answer: ['B', 'C', 'D']