A) \[\frac{e}{2}\]
B) \[\frac{1}{2}\]
C) \[\frac{e-2}{2}\]
D) None
Correct Answer: C
Solution :
[c] \[I=\int_{1}^{e}{{{\left\{ \frac{(logx-1)}{1+{{(logx)}^{2}}} \right\}}^{2}}}dx\] put \[log\,x=t\text{ }\Rightarrow x={{e}^{t}}\] \[dx={{e}^{t}}dt\] \[I={{\int_{0}^{1}{{{e}^{t}}\left\{ \frac{t-1}{1+{{t}^{2}}} \right\}}}^{2}}dt\] \[=\int_{0}^{1}{{{e}^{t}}\left\{ \frac{t-1}{1+{{t}^{2}}}-\frac{2t}{{{(1+{{t}^{2}})}^{2}}} \right\}}dt\] \[=\left[ \frac{{{e}^{t}}}{1+{{t}^{2}}} \right]_{0}^{1}=\frac{e}{2}-1=\frac{e-2}{2}\]
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