Question

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Answer 1

$\displaystyle \int_{e^n}^{e^{n+1}}\dfrac{\ln^2 x}{x}\, dx = \dfrac{1}{3}\int_{e^n}^{e^{n+1}}3\ln^2 x\cdot (\ln x)'\, dx = \\ \\ = \dfrac{1}{3}\cdot \ln^3 x\Big|_{e^n}^{e^{n+1}} = \dfrac{\ln^3(e^{n+1}) -\ln^3(e^n)}{3} = \dfrac{(n+1)^3-n^3}{3} = \dfrac{7}{3} \\ \\ \Rightarrow (n+1-n)((n+1)^2+(n+1)n+n^2)= 7 \\ \\ n^2+2n+1+n^2+n+n^2 = 7 \\3n^2+3n-6 = 0 \\ n^2+n-2 = 0\\ \\ \Delta = 1+8 = 9 \Rightarrow n_{1,2} = \dfrac{-1\pm 3}{2}\Rightarrow \boxed{n = 1}$