integral from 0 to 4 of (sqrt(x))/(1+x)

解

\int_{0}^{4} \frac{x}{1 + x} d x

2 (- arctan (2) + 2)

十進法表記

1.78570 \dots

解答ステップ

\int_{0}^{4} \frac{x}{1 + x} d x

u置換積分法を適用する

= \int_{0}^{2} \frac{2 u ^{2}}{1 + u ^{2}} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2} \frac{u ^{2}}{1 + u ^{2}} d u

\frac{u ^{2}}{1 + u ^{2}} = - \frac{1}{1 + u ^{2}} + 1

= 2 \cdot \int_{0}^{2} - \frac{1}{1 + u ^{2}} + 1 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

= 2 (- \int_{0}^{2} \frac{1}{1 + u ^{2}} d u + \int_{0}^{2} 1 d u)

\int_{0}^{2} \frac{1}{1 + u ^{2}} d u = arctan (2)

\int_{0}^{2} 1 d u = 2

= 2 (- arctan (2) + 2)