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$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\frac{1}{2}(g(x))^2+C$, (where $C$ is constant of integration.) Then $g(x)=$
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The correct answer is:
$\log \left(x+\sqrt{1+x^2}\right)$
$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\int t \mathrm{~d} t=\frac{t^2}{2}+C$
[where $t=\log \left(x+\sqrt{1+x^2}\right)$ ]
$\begin{aligned}
& =\frac{\left\{\log \left(x+\sqrt{1+x^2}\right)\right\}^2}{2}+C \\
& \Rightarrow g(x)=\log \left(x+\sqrt{1+x^2}\right)^2
\end{aligned}$
[where $t=\log \left(x+\sqrt{1+x^2}\right)$ ]
$\begin{aligned}
& =\frac{\left\{\log \left(x+\sqrt{1+x^2}\right)\right\}^2}{2}+C \\
& \Rightarrow g(x)=\log \left(x+\sqrt{1+x^2}\right)^2
\end{aligned}$
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