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) \[\sqrt{1+{{x}^{2}}}+\log (\sec x+\tan x)+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{x+1}{\sqrt{{{x}^{2}}+1}}\,dx=\int_{{}}^{{}}{\frac{x}{\sqrt{{{x}^{2}}+1}}\,dx+\int_{{}}^{{}}{\frac{1}{\sqrt{{{x}^{2}}+1}}\,dx}}}\] Put \[{{x}^{2}}+1=t\Rightarrow 2x\,dx=dt\], then it reduce to \[\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{{{t}^{1/2}}}+\int_{{}}^{{}}{\frac{1}{\sqrt{{{x}^{2}}+1}}}}\,dx=\frac{1}{2}.2.{{t}^{1/2}}+\log (x+\sqrt{{{x}^{2}}+1})+c\] \[={{({{x}^{2}}+1)}^{1/2}}+\log (x+\sqrt{{{x}^{2}}+1})+c.\]
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