A) \[\frac{1}{2\sqrt{2}}\log \left\{ \frac{\sqrt{1+{{x}^{2}}}+x\sqrt{2}}{\sqrt{1+{{x}^{2}}}-x\sqrt{2}} \right\}+c\]
B) \[\frac{1}{2\sqrt{2}}\log \left\{ \frac{\sqrt{1+{{x}^{2}}}-\sqrt{2}}{\sqrt{1+{{x}^{2}}}+\sqrt{2}} \right\}+c\]
C) \[\frac{1}{2\sqrt{2}}\log \left\{ \frac{\sqrt{1+{{x}^{2}}}-x\sqrt{2}}{\sqrt{1+{{x}^{2}}}+x\sqrt{2}} \right\}+c\]
D) None of these
Correct Answer: C
Solution :
Put \[x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta \,d\theta ,\] then \[\int_{{}}^{{}}{\frac{dx}{({{x}^{2}}-1)\sqrt{{{x}^{2}}+1}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\theta \,d\theta }{({{\tan }^{2}}\theta -1)\sec \theta }}}=\int_{{}}^{{}}{\frac{\cos \theta \,d\theta }{(2{{\sin }^{2}}\theta -1)}}\] Again put \[t=\sin \theta \Rightarrow dt=\cos \theta \,d\theta ,\] then it reduces to \[\int_{{}}^{{}}{\frac{dt}{(2{{t}^{2}}-1)}}=\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}-{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}}}=\frac{1}{2\sqrt{2}}\log \left( \frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}} \right)+c\] \[=\frac{1}{2\sqrt{2}}\log \left( \frac{\sqrt{1+{{x}^{2}}}-x\sqrt{2}}{\sqrt{1+{{x}^{2}}}+x\sqrt{2}} \right)+c\].
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