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

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

Vedclass · Answer

(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 $I = \int \left[ \frac{t - 1}{1 + t^2} \right]^2 e^t dt = \int e^t \left[ \frac{t^2 - 2t + 1}{(1 + t^2)^2} \right] dt$. We can rewrite the numerator as $(t^2 + 1) - 2t$. So,$I = \int e^t \left[ \frac{t^2 + 1}{(1 + t^2)^2} - \frac{2t}{(1 + t^2)^2} \right] dt = \int e^t \left[ \frac{1}{1 + t^2} - \frac{2t}{(1 + t^2)^2} \right] dt$. This is in the form $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \frac{1}{1 + t^2}$. Thus,$I = e^t \left( \frac{1}{1 + t^2} \right) + c$. Substituting back $t = \log x$,we get $I = \frac{x}{1 + (\log x)^2} + c$.

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

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