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$f(x)=\frac{1+2x \sqrt{x^2+1}+2x^2 }{1+x+ \sqrt{x^2+1} } =\frac{1+2x \sqrt{x^2+1}+2x^2 }{1+x+ \sqrt{x^2+1} }$
$=\frac{(1+2x \sqrt{x^2+1}+2x^2 )(1+x-\sqrt{x^2+1})}{2x }$
$=\frac{1}{2}\left[ {\sqrt{x^2+1}+x+\frac{(x-1)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}}} \right]+\frac{x}{2\sqrt{x^2+1}(\sqrt{x^2+1}+1)}+\frac{1}{2\sqrt{x^2+1}}$
Do đó
$\int f(x)dx=\frac{1}{2}\left[ {(x-1)(\sqrt{x^2+1}+x)+\ln(\sqrt{x^2+1}+1)+\ln(\sqrt{x^2+1}+x)} \right]+C$
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