A) \[\log (1+\sqrt{2})\]
B) \[\log (1+\sqrt{2})-\frac{\pi }{4}\]
C) \[\log (1+\sqrt{2})+\frac{\pi }{2}\]
D) None of these
Correct Answer: B
Solution :
\[f(x)=\int{\frac{{{x}^{2}}dx}{(1+{{x}^{2}})\left( 1+\sqrt{1+{{x}^{2}}} \right)}}\] Let \[x=\tan \theta ,dx={{\sec }^{2}}\theta \,d\theta \]\[=(1+{{x}^{2}}).d\theta \] \[f(x)=\int{\frac{{{x}^{2}}dx}{(1+{{x}^{2}})\left( 1+\sqrt{1+{{x}^{2}}} \right)}}=\int{\frac{{{\tan }^{2}}\theta {{\sec }^{2}}\theta d\theta }{{{\sec }^{2}}\theta (1+\sec \theta )}}\] \[=\int{\frac{{{\tan }^{2}}\theta \,\,d\theta }{1+\sec \theta }\theta }=\int{\frac{{{\sin }^{2}}\theta \,d\theta }{\cos \theta (1+\cos \theta )}}=\int{\frac{1-{{\cos }^{2}}\theta \,d\theta }{\cos \theta (1+\cos \theta )}}\] \[=\int{\frac{(1-\cos \theta )d.\theta }{\cos \theta }}\]\[=\int{\sec \theta d\theta -\int{d\theta }}\] \[=\log (x+\sqrt{1+{{x}^{2}}})-{{\tan }^{-1}}x+c\] \[f(0)=\log (0+\sqrt{1+0}-{{\tan }^{-1}}(0)+c\] \[0=\log 1-0+c\] Þ \[c=0\] \[f(1)=\log (1+\sqrt{1+{{1}^{2}}})-{{\tan }^{-1}}(1)=\log (1+\sqrt{2})-\frac{\pi }{4}\].
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