Question

integral 0to1 x/[1+√(1+x*2)] dx​

Answer 1

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Answer 2

before we start i wanna tell u that for me

$f = > \: integral$

therefore,

$f \frac{x.dx}{1 + \sqrt{1 + {x}^{2} } } \\ put \: 1 + {x }^{2} = {t}^{2} \\ therefore \: x.dx = t.dt \\ so \\ f \frac{t.dt}{1 + t} = fdt \: - f \frac{dt}{1 + t} \\ = t - log(1 + t) + c \\ on \: putting \: value \: of \: t \: we \: get \\ = \frac{1}{0} ( \sqrt{1 + {x}^{2} } - log(1 + \sqrt{1 + {x}^{2} } ) \\ = (( \sqrt{1 + {1}^{2} } - log(1 + \sqrt{1 + {1}^{2} } ) ) - (\sqrt{1 + {0}^{2} } - log(1 + \sqrt{1 + {0}^{2} }))) \\ = ( \sqrt{2 } - log(1 + \sqrt{2} ) ) - (1 - log( \sqrt{2} ) ) \\ = \sqrt{2} - 1 + log( \frac{ \sqrt{2} }{1 + \sqrt{2} } )$