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$\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x=$
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The correct answer is:
$\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C$
Let $I=\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x$
$\begin{aligned} & \text { put } x+\sqrt{1+x^2}=t \\ & \left(1+\frac{x}{\sqrt{1+x^2}}\right) d x=d t \Rightarrow\left(\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right) d x=d t \\ & \because \quad I=\int t d t=\frac{t^2}{2}+C=\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C\end{aligned}$
$\begin{aligned} & \text { put } x+\sqrt{1+x^2}=t \\ & \left(1+\frac{x}{\sqrt{1+x^2}}\right) d x=d t \Rightarrow\left(\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right) d x=d t \\ & \because \quad I=\int t d t=\frac{t^2}{2}+C=\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C\end{aligned}$
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