A) \[\frac{\pi }{4}-\frac{1}{2}\log 2\]
B) \[\pi -\frac{1}{2}\log 2\]
C) \[\frac{\pi }{4}-\log 2\]
D) \[\pi -\log 2\]
Correct Answer: A
Solution :
Put \[x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta \,\,d\theta \] Also as \[x=0,\theta =0\]and \[x=1,\theta =\frac{\pi }{4}\] Therefore, \[\int_{0}^{1}{{{\tan }^{-1}}x\,dx=\int_{0}^{\pi /4}{\theta {{\sec }^{2}}\theta \,d\theta }}\] \[=\frac{\pi }{4}\]\[-\log \sqrt{2}=\frac{\pi }{4}-\frac{1}{2}\log 2\].
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