A) \[x{{\tan }^{-1}}x+c\]
B) \[x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c\]
C) \[2x{{\tan }^{-1}}x+\log (1+{{x}^{2}})+c\]
D) \[2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c\]
Correct Answer: D
Solution :
Put \[x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta \,d\theta ,\] then \[\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}\,dx}=\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }}\,\,\,{{\sec }^{2}}\theta \,d\theta \] \[=\int_{{}}^{{}}{{{\tan }^{-1}}(\tan 2\theta ){{\sec }^{2}}\theta \,d\theta }=\int_{{}}^{{}}{2\theta {{\sec }^{2}}\theta \,d\theta }\] \[=2\left[ \theta \tan \theta -\int_{{}}^{{}}{\tan \theta \,d\theta } \right]\]\[=2x{{\tan }^{-1}}x-\log ({{x}^{2}}+1)+c.\]
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