The value of int_1^{e^2} frac{dx}{x(1 + ln x) | vedclass

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(A) Let $I = \int_1^{e^2} \frac{dx}{x(1 + \ln x)^2}$.Substitute $t = 1 + \ln x$.Then,$dt = \frac{1}{x} dx$.Change the limits of integration:When $x =

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$2/3$

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(A) Let $I = \int_1^{e^2} \frac{dx}{x(1 + \ln x)^2}$.Substitute $t = 1 + \ln x$.Then,$dt = \frac{1}{x} dx$.Change the limits of integration:When $x = 1$,$t = 1 + \ln(1) = 1 + 0 = 1$.When $x = e^2$,$t = 1 + \ln(e^2) = 1 + 2 = 3$.Now,substitute these into the integral:$I = \int_1^3 \frac{dt}{t^2} = \int_1^3 t^{-2} dt$.Integrating $t^{-2}$ gives $\left[ \frac{t^{-1}}{-1} \right]_1^3 = \left[ -\frac{1}{t} \right]_1^3$.Evaluating at the limits:$I = -\left( \frac{1}{3} - \frac{1}{1} \right) = -\left( \frac{1}{3} - 1 \right) = -\left( -\frac{2}{3} \right) = \frac{2}{3}$.

The value of $\int_1^{e^2} \frac{dx}{x(1 + \ln x)^2}$ is

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