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Question Number 195910 by Calculusboy last updated on 13/Aug/23

	Answered by witcher3 last updated on 20/Aug/23

	$Methode of differentiation ?$ $A = \ln (2) \int_{0}^{1} \frac{x^{2} - 1}{\ln (x)} dx$ $A (a) = \ln (2) \int_{0}^{1} \frac{x^{a} - 1}{\ln (x)} dx, a ⩾ 0$ $x \to \frac{x^{a} - 1}{\ln (x)}, x \to 1, a > 0$ $= \frac{x - 1}{\ln (x)} . \frac{x^{a} - 1}{x - 1}, \frac{x - 1}{\ln (x)} \to 1$ $\frac{x^{a} - 1}{x - 1} \to a \Rightarrow \frac{x^{a} - 1}{\ln (x)} \to a, integrable near 1, near 0$ $\to 0 also its riemann integrabl + continuity \Rightarrow$ $uniformaly continue$ $\forall x \in [0, 1] x \to \frac{x^{a} - 1}{\ln (x)} is differwntiable as function$ $of variable a \Rightarrow we can switch \int and \partial in E$ $A (0) = 0, A = \int_{0}^{2} A^{'} da$ $A^{'} = \ln (2) \int_{0}^{1} \partial_{a} \frac{e^{aln (x)} - 1}{\ln (x)} dx = \int_{0}^{1} x^{a} dx = \frac{1}{a + 1} . . E$ $A = \ln (2) \int_{0}^{2} \frac{1}{a + 1} da = \ln (2) \ln (3)$
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