int_1^{e^2} frac{dx}{x(1+log x)^2} = | vedclass

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Question

(A) Let $I = \int_1^{e^2} \frac{dx}{x(1+\log x)^2}$.Substitute $t = 1 + \log x$. Then $dt = \frac{1}{x} dx$.When $x = 1$,$t = 1 + \log(1) = 1 + 0 = 1$

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) Let $I = \int_1^{e^2} \frac{dx}{x(1+\log x)^2}$.Substitute $t = 1 + \log x$. Then $dt = \frac{1}{x} dx$.When $x = 1$,$t = 1 + \log(1) = 1 + 0 = 1$.When $x = e^2$,$t = 1 + \log(e^2) = 1 + 2 = 3$.Thus,$I = \int_1^3 \frac{dt}{t^2} = \int_1^3 t^{-2} dt$.Integrating,we get $I = \left[ \frac{t^{-1}}{-1} \right]_1^3 = \left[ -\frac{1}{t} \right]_1^3$.Evaluating the limits,$I = \left( -\frac{1}{3} \right) - \left( -\frac{1}{1} \right) = -\frac{1}{3} + 1 = \frac{2}{3}$.

$\int_1^{e^2} \frac{dx}{x(1+\log x)^2} = $

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