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

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

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

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(B) Let $I = \int \left\{ \frac{\log x - 1}{1 + (\log x)^2} \right\}^2 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. The integral becomes $I = \int e^t \left( \frac{t - 1}{1 + t^2} \right)^2 dt$. Wait,the original expression in the prompt is $\int \left\{ \frac{\log x - 1}{1 + (\log x)^2} \right\}^2 dx$. However,based on the standard form $\int e^t [f(t) + f'(t)] dt$,the integral is likely $\int \frac{e^t (t-1)}{(1+t^2)^2} dt$ or similar. Given the provided solution steps: Let $f(t) = \frac{1}{1+t^2}$. Then $f'(t) = \frac{-2t}{(1+t^2)^2}$. Thus,$\int e^t [f(t) + f'(t)] dt = e^t f(t) + C = \frac{e^t}{1+t^2} + C$. Substituting $t = \log x$ and $e^t = x$,we get $\frac{x}{1 + (\log x)^2} + C$.

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

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