A) \[\frac{\log \,x}{{{(log\,x)}^{2}}+1}+c\]
B) \[\frac{x}{{{x}^{2}}+1}+c\]
C) \[\frac{x{{e}^{x}}}{1+{{x}^{2}}}+c\]
D) \[\frac{x}{{{(log\,x)}^{2}}+1}+c\]
Correct Answer: D
Solution :
[d] \[\int{\frac{{{(\log x-1)}^{2}}}{{{(1+{{(\log x)}^{2}})}^{2}}}dx}\] \[=\int{\left[ \frac{1}{(1+{{(\log x)}^{2}})}-\frac{2\log x}{{{(1+{{(\log x)}^{2}})}^{2}}} \right]}dx\] \[=\int{\left[ \frac{{{e}^{t}}}{1+{{t}^{2}}}-\frac{2t{{e}^{t}}}{{{(1+{{t}^{2}})}^{2}}} \right]}dt\] (Putting\[\log x=t\Rightarrow dx={{e}^{t}}dt\]) \[\int{{{e}^{t}}\left[ \frac{1}{1+{{t}^{2}}}-\frac{2t}{{{(1+{{t}^{2}})}^{2}}} \right]dt}\] \[=\frac{{{e}^{t}}}{1+{{t}^{2}}}+c\] \[=\frac{x}{1+{{(\log x)}^{2}}}+c\] (Resubstituting \[t=\log x\])You need to login to perform this action.
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