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

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

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

Vedclass · Answer

(C) Let $I = \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} dx$. Substitute $\log x = t$,so $x = e^t$ and $dx = e^t dt$. Then $I = \int \frac{(t-1)^2}{(1+t^2)^2} e^t dt = \int e^t \left( \frac{t^2-2t+1}{(1+t^2)^2} \right) dt$. We can rewrite the integrand as $\int e^t \left( \frac{t^2+1-2t}{(1+t^2)^2} \right) dt = \int e^t \left( \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right) dt$. 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,$I = e^t \left( \frac{1}{1+t^2} \right) + C$. Substituting back $t = \log x$,we get $I = \frac{e^{\log x}}{1+(\log x)^2} + C = \frac{x}{1+(\log x)^2} + C$.

$\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=$ (where $C$ is constant of integration.)

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