question_answer

\[{{\int{\left\{ \frac{(\log x-1)}{1+{{(\log \,x)}^{2}}} \right\}}}^{2}}dx\]is equal to     AIEEE  Solved  Paper-2005

A) \[\frac{x}{{{(\log \,x)}^{2}}+1}+C\]

B) \[\frac{x{{e}^{x}}}{1+\,{{x}^{2}}}+C\]

C) \[\frac{x}{\,{{x}^{2}}+1}+C\]

D) \[\frac{\log \,\,x}{\,{{(\log x)}^{2}}+1}+C\]

Correct Answer: A

Solution :

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