integral from 0 to 1 of sqrt(2/(1+x))

解

\int_{0}^{1} \frac{2}{1 + x} d x

4 - 22

十進法表記

1.17157 \dots

解答ステップ

\int_{0}^{1} \frac{2}{1 + x} d x

部分積分を適用する

= [\frac{2}{1 + x} x - \int - \frac{x}{2 ( x + 1 ) ^{\frac{3}{2}}} d x]_{0}^{1}

\int - \frac{x}{2 ( x + 1 ) ^{\frac{3}{2}}} d x = - \frac{2 ( x + 2 )}{x + 1}

= [\frac{2}{1 + x} x - (- \frac{2 ( x + 2 )}{x + 1})]_{0}^{1}

簡素化

[\frac{2}{1 + x} x - (- \frac{2 ( x + 2 )}{x + 1})]_{0}^{1} : [22 1 + x]_{0}^{1}

= [22 1 + x]_{0}^{1}

限度を計算する:

4 - 22

= 4 - 22