A) \[\log \,(\sqrt{2}+1)\]
B) \[\log \,(\sqrt{2}+1)-\frac{\pi }{4}\]
C) \[\log \,(\sqrt{2}+1)+\frac{\pi }{4}\]
D) None of these
Correct Answer: B
Solution :
| We have, \[f(x)=\int{\frac{{{x}^{2}}}{(1+{{x}^{2}})(1+\sqrt{1+{{x}^{2}}})}dx}\] |
| \[\Rightarrow \] \[f(x)=\int{\frac{{{x}^{2}}(\sqrt{1+{{x}^{2}}}-1)}{(1+{{x}^{2}})(1+{{x}^{2}}-1)}dx}\] |
| \[\Rightarrow \] \[f(x)=\int{\frac{\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}dx-\int{\frac{1}{1+{{x}^{2}}}}dx}\] |
| \[\Rightarrow \] \[f(x)=\int{\frac{1}{\sqrt{1+{{x}^{2}}}}dx-\int{\frac{1}{1+{{x}^{2}}}dx}}\] |
| \[\Rightarrow \] \[f(x)=\log (x+\sqrt{1+{{x}^{2}})}-{{\tan }^{-1}}x+C\] |
| \[\Rightarrow \] \[f(0)=0=C\] |
| \[\therefore \] \[f(x)=\log \,(x+\sqrt{1+{{x}^{2}}})-{{\tan }^{-1}}x\] |
| \[\Rightarrow \] \[f(1)=\log \,(1+\sqrt{2})-{{\tan }^{-1}}1\] |
| \[f(1)=\log \,(\sqrt{2}+1)-\frac{\pi }{4}\] |
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