int left( frac{1-log x}{1+(log x)^2} right)^2 | vedclass

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(C) Let $I = \int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \in

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$\frac{x}{1+(\log x)^2}+c$

Vedclass · Answer

(C) Let $I = \int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int \left( \frac{1-t}{1+t^2} \right)^2 e^t dt$. This is of the form $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$. Let $f(t) = \frac{t}{1+t^2}$. Then $f'(t) = \frac{(1+t^2)(1) - t(2t)}{(1+t^2)^2} = \frac{1-t^2}{(1+t^2)^2}$. This does not match directly. Let us rewrite the integral: $I = \int e^t \frac{(1-t)^2}{(1+t^2)^2} dt = \int e^t \frac{1-2t+t^2}{(1+t^2)^2} dt = \int e^t \left( \frac{1+t^2-2t}{(1+t^2)^2} \right) dt = \int e^t \left( \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right) dt$. Here,if $f(t) = \frac{1}{1+t^2}$,then $f'(t) = \frac{-2t}{(1+t^2)^2}$. Thus,$I = e^t \left( \frac{1}{1+t^2} \right) + c$. Substituting back $t = \log x$,we get $I = \frac{e^{\log x}}{1+(\log x)^2} + c = \frac{x}{1+(\log x)^2} + c$.

$\int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx = $

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