Let \[I = \int_0^1 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx\]

Let’s put \[ 1 + {x^2} = {t^2}\]

\[ \Rightarrow \] \[2xdx = 2tdt\]

\[ \Rightarrow \] \[ xdx = tdt\]

therefore,\[ I = \int_1^{\sqrt 2 } {\frac{{tdt}}{t}} \]

\[ = [t]_1^{\sqrt 2 } = \sqrt 2 – 1\]

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