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

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(B) Let $I = \int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx$.Put $\log x = t$,then $x = e^{t}$,so $dx = e^{t} dt$.Substituting these into th

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

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(B) Let $I = \int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx$.Put $\log x = t$,then $x = e^{t}$,so $dx = e^{t} dt$.Substituting these into the integral,we get $I = \int e^{t} \left[\frac{1-t}{1+t^{2}}\right]^{2} dt$.Expanding the square,$I = \int e^{t} \frac{1+t^{2}-2t}{(1+t^{2})^{2}} dt = \int e^{t} \left[\frac{1}{1+t^{2}} - \frac{2t}{(1+t^{2})^{2}}\right] dt$.Using the standard integral 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,$I = e^{t} \left(\frac{1}{1+t^{2}}\right) + c$.Substituting back $t = \log x$,we get $I = \frac{x}{1+(\log x)^{2}} + c$.

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

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