$\int_0^1 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx$
$\int_0^1 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx$
Official Solution
VVidaara Team
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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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