A) \[\frac{1}{2}{{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c\]
B) \[\log {{(x+\sqrt{1+{{x}^{2}}})}^{2}}+c\]
C) \[\log (x+\sqrt{1+{{x}^{2}}})+c\]
D) None of these
Correct Answer: A
Solution :
Put \[\log (x+\sqrt{1+{{x}^{2}}})=t\Rightarrow \frac{1}{\sqrt{1+{{x}^{2}}}}\,dx=dt,\] then \[\int_{{}}^{{}}{\frac{\log (x+\sqrt{1+{{x}^{2}}})}{\sqrt{1+{{x}^{2}}}}\,dx}=\int_{{}}^{{}}{t\,dt}\] \[=\frac{1}{2}{{\left[ \log (x+\sqrt{1+{{x}^{2}}}) \right]}^{2}}+c\].You need to login to perform this action.
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