A) a
B) \[\frac{a}{2}\]
C) \[a\sqrt{2}\]
D) \[\frac{a}{\sqrt{2}}\]
Correct Answer: A
Solution :
\[I=\int\limits_{0}^{1}{1.}\sqrt{|\ln x|}dx=\left[ x\sqrt{|\ln x|} \right]_{0}^{1}-\int_{{}}^{{}}{\frac{x}{2\sqrt{|\ln x|}}}\left( -\frac{1}{x} \right)dx\]\[=\frac{1}{2}\int\limits_{0}^{1}{\frac{1}{\sqrt{|\ln x|}}}dx\] Put\[|\ln x|={{t}^{2}}\] \[\Rightarrow \]\[l=\frac{1}{2}I\int\limits_{0}^{\infty }{\frac{1}{t}}\left( 2t{{e}^{-{{t}^{2}}}} \right)dt=\int\limits_{0}^{\infty }{{{e}^{-{{x}^{2}}}}}dx=a\]
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