MCQ
${\int {\left\{ {\frac{{(\log x - 1)}}{{1 + {{(\log x)}^2}}}} \right\}} ^2}dx$=
- A$\frac{{x{e^x}}}{{1 + {x^2}}} + c$
- ✓$\frac{x}{{{{(\log x)}^2} + 1}} + C$
- C$\frac{{\log x}}{{{{(\log x)}^2} + 1}} + c$
- D$\frac{x}{{{x^2} + 1}} + c$
Put $\log x = t \Rightarrow dx = {e^t}dt$
Integral $ = \int {{e^t}\left[ {\frac{1}{{1 + {t^2}}} - \frac{{2t}}{{{{(1 + {t^2})}^2}}}} \right]} \;dt$
$\left[ {\because \;\int {{e^x}[f(x) + f'(x)]\;dx = {e^x}f(x) + c} \;} \right]$
$ = \frac{{{e^t}}}{{1 + {t^2}}} + C = \frac{x}{{1 + {{(\log x)}^2}}} + C$.
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