A) \[\frac{x{{e}^{x}}}{1+{{x}^{2}}}+c\]
B) \[\frac{x}{{{(\log x)}^{2}}+1}+C\]
C) \[\frac{\log x}{{{(\log x)}^{2}}+1}+c\]
D) \[\frac{x}{{{x}^{2}}+1}+c\]
Correct Answer: B
Solution :
\[{{\int{\left\{ \frac{\log x-1}{1+{{(\log x)}^{2}}} \right\}}}^{2}}dx\]. 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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