\[\int_{{}}^{{}}{\frac{x+1}{\sqrt{1+{{x}^{2}}}}dx}=\] [MP PET 1991]

\[\sqrt{1+{{x}^{2}}}+\log (\sec x+\tan x)+c\]

\[\sqrt{1+{{x}^{2}}}+{{\tan }^{-1}}x+c\]

\[\sqrt{1+{{x}^{2}}}-\log \{x+\sqrt{1+{{x}^{2}}}\}+c\]

\[\sqrt{1+{{x}^{2}}}+\log \{x+\sqrt{1+{{x}^{2}}}\}+c\]

\[\int_{{}}^{{}}{\frac{x+1}{\sqrt{1+{{x}^{2}}}}dx}=\] [M

1.	\[\int_{{}}^{{}}{\frac{x+1}{\sqrt{1+{{x}^{2}}}}dx}=\] [MP PET 1991]
A.	\[\sqrt{1+{{x}^{2}}}+{{\tan }^{-1}}x+c\]
B.	\[\sqrt{1+{{x}^{2}}}-\log \{x+\sqrt{1+{{x}^{2}}}\}+c\]
C.	\[\sqrt{1+{{x}^{2}}}+\log \{x+\sqrt{1+{{x}^{2}}}\}+c\]
D.	\[\sqrt{1+{{x}^{2}}}+\log (\sec x+\tan x)+c\]
Answer» D. \[\sqrt{1+{{x}^{2}}}+\log (\sec x+\tan x)+c\]

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\[\int_{{}}^{{}}{\frac{x+1}{\sqrt{1+{{x}^{2}}}}dx}=\] [MP PET 1991]

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