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

解

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

2 - ln (3)

十進法表記

0.90138 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{3} \frac{u - 1}{u} d u

拡張

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int_{1}^{3} 1 - \frac{1}{u} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} 1 d u - \int_{1}^{3} \frac{1}{u} d u

\int_{1}^{3} 1 d u = 2

\int_{1}^{3} \frac{1}{u} d u = ln (3)

= 2 - ln (3)