A) \[\frac{x+{{\tan }^{-1}}x}{\sqrt{1+{{x}^{2}}}}+c\]
B) \[\frac{x-{{\tan }^{-1}}x}{\sqrt{1+{{x}^{2}}}}+c\]
C) \[\frac{{{\tan }^{-1}}x-x}{\sqrt{1+{{x}^{2}}}}+c\]
D) None of these
Correct Answer: B
Solution :
Put \[x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta \,d\theta ,\] then \[\int_{{}}^{{}}{\frac{x{{\tan }^{-1}}x}{{{(1+{{x}^{2}})}^{3/2}}}\,dx}=\int_{{}}^{{}}{\frac{\theta \tan \theta {{\sec }^{2}}\theta \,d\theta }{{{(1+{{\tan }^{2}}\theta )}^{3/2}}}}\] \[=\int_{{}}^{{}}{\theta \sin \theta \,d\theta }=-\theta \cos \theta +\sin \theta +c\] \[=\frac{x}{\sqrt{{{x}^{2}}+1}}-{{\tan }^{-1}}x\frac{1}{\sqrt{{{x}^{2}}+1}}=\frac{x-{{\tan }^{-1}}x}{\sqrt{1+{{x}^{2}}}}+c\].
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