A) \[\frac{x-{{\tan }^{-1}}x}{1-{{x}^{2}}}+c\]
B) \[\frac{x+{{\tan }^{-1}}x}{\sqrt{1-{{x}^{2}}}}+c\]
C) \[\frac{x-{{\tan }^{-1}}x}{\sqrt{1+{{x}^{2}}}}+c\]
D) \[\frac{x+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}}+c\]
Correct Answer: C
Solution :
Let\[I=\int{\frac{x{{\tan }^{-1}}x}{{{(1+{{x}^{2}})}^{3/2}}}}dx\] Put\[x\tan \theta \Rightarrow dx={{\sec }^{2}}\theta d\theta \] \[\therefore \] \[I=\int{\frac{\tan \theta {{\tan }^{-1}}(\tan \theta )}{{{(1+{{\tan }^{2}}\theta )}^{3/2}}}{{\sec }^{2}}\theta d\theta }\] \[\Rightarrow \] \[I=\int{\frac{\theta \tan \theta {{\sec }^{2}}\theta }{{{\sec }^{3}}\theta }d\theta }\] \[\Rightarrow \] \[I=\int{\theta \sin \theta \,\,d\theta }\] \[\Rightarrow \] \[I=\theta (-\cos \theta )+\int{\cos \theta \,\,d\theta }\] \[\Rightarrow \,\,\,\,I=\theta (-\cos \theta )\,+\sin \theta +c\] \[\Rightarrow \] \[I=-{{\tan }^{-1}}x\frac{1}{\sqrt{1+{{x}^{2}}}}+\frac{x}{\sqrt{1+{{x}^{2}}}}+c\] \[\Rightarrow \,\,\,\,\,\,\,\,I=\frac{(x-{{\tan }^{-1}}x)}{\sqrt{1+{{x}^{2}}}}+c\]
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