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$$ \int \sqrt{1 + {1 \over t^2} + {2 \over t}}\,\mathrm dt$$

I tried making substitution, using $ u=1 + \dfrac{1}{ t^2} + \dfrac{2 }{ t} $, then , $dt=\dfrac{du}{-2\left({1 \over t^3 }+ {1 \over t}\right)}$, which does not help, because the $t$'s are not eliminated...

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Use the following :
$$ a^2 + 2ab + b^2 = (a+b)^2$$

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$$ \int \sqrt{1 + {1 \over t^2} + {2 \over t}} dt=\int {1\over t}\sqrt{t^2+2t+1} \,\,\,dt$$

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Hints:

$$ 1 + {1 \over t^2} + {2 \over t}=\frac{(t+1)^2}{t^2}$$

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