A) \[{{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c\]
B) \[x\log (x+\sqrt{1+{{x}^{2}}})+c\]
C) \[\frac{1}{2}\log (x+\sqrt{1+{{x}^{2}}})+c\]
D) \[\frac{1}{2}{{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c\]
E) \[\frac{x}{2}\log (x+\sqrt{1+{{x}^{2}}})+c\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{\log (x+\sqrt{1+{{x}^{2}}})}{\sqrt{1+{{x}^{2}}}}}dx\] Put \[t=\log (x+\sqrt{1+{{x}^{2}}})\] \[\frac{dt}{dx}=\frac{1}{x+\sqrt{1+{{x}^{2}}}}\left\{ 1+\frac{x}{\sqrt{1+{{x}^{2}}}} \right\}\] \[=\frac{x+\sqrt{1+{{x}^{2}}}}{(x+\sqrt{1+{{x}^{2}}}).\sqrt{1+{{x}^{2}}}}\] \[=\frac{1}{\sqrt{1+{{x}^{2}}}}\] \[\therefore \] \[\frac{dx}{\sqrt{1+{{x}^{2}}}}=dt\] \[\therefore \] \[I=\int{t}\,dt=\frac{{{t}^{2}}}{2}+c\] \[=\frac{1}{2}{{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c\]
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