If f(x) = frac{|log x|}{x^2},then int_{1/e}^e | vedclass

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(D) We have $f(x) = \frac{|\log x|}{x^2} = \begin{cases} -\frac{\log x}{x^2}, & \frac{1}{e} \le x < 1 \ \frac{\log x}{x^2}, & 1 \le x \le e \end{case

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$2(1 - \frac{1}{e})$

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(D) We have $f(x) = \frac{|\log x|}{x^2} = \begin{cases} -\frac{\log x}{x^2}, & \frac{1}{e} \le x We split the integral at $x = 1$:$\int_{1/e}^e f(x) dx = \int_{1/e}^1 -\frac{\log x}{x^2} dx + \int_1^e \frac{\log x}{x^2} dx$.Using integration by parts for $\int \frac{\log x}{x^2} dx$,let $u = \log x$ and $dv = x^{-2} dx$. Then $du = \frac{1}{x} dx$ and $v = -\frac{1}{x}$.$\int \frac{\log x}{x^2} dx = -\frac{\log x}{x} - \int -\frac{1}{x^2} dx = -\frac{\log x}{x} - \frac{1}{x} + C$.Evaluating the definite integrals:$\int_{1/e}^1 -\frac{\log x}{x^2} dx = -\left[ -\frac{\log x}{x} - \frac{1}{x} \right]_{1/e}^1 = \left[ \frac{\log x + 1}{x} \right]_{1/e}^1 = (\frac{0+1}{1}) - (\frac{\log(1/e) + 1}{1/e}) = 1 - ((-1+1) \cdot e) = 1 - 0 = 1$.$\int_1^e \frac{\log x}{x^2} dx = \left[ -\frac{\log x + 1}{x} \right]_1^e = (-\frac{\log e + 1}{e}) - (-\frac{\log 1 + 1}{1}) = -\frac{2}{e} + 1 = 1 - \frac{2}{e}$.Summing the parts: $1 + (1 - \frac{2}{e}) = 2 - \frac{2}{e} = 2(1 - \frac{1}{e})$.

If $f(x) = \frac{|\log x|}{x^2}$,then $\int_{1/e}^e f(x) dx =$

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