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

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

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

Vedclass · Answer

(C) Let $I = \int \left( \frac{\log x - 1}{1 + (\log x)^2} \right)^2 dx$. Substitute $t = \log x$,then $x = e^t$ and $dx = e^t dt$. $I = \int e^t \frac{(t - 1)^2}{(1 + t^2)^2} dt = \int e^t \frac{t^2 - 2t + 1}{(1 + t^2)^2} dt$. $I = \int e^t \left( \frac{t^2 + 1 - 2t}{(1 + t^2)^2} \right) dt = \int e^t \left( \frac{1}{1 + t^2} - \frac{2t}{(1 + t^2)^2} \right) dt$. Using the formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$,where $f(t) = \frac{1}{1 + t^2}$ and $f'(t) = -\frac{2t}{(1 + t^2)^2}$. Thus,$I = e^t \left( \frac{1}{1 + t^2} \right) + C = \frac{x}{1 + (\log x)^2} + C$.

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

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