integral from 1 to 5 of (sqrt(u-1))/u

解

\int_{1}^{5} \frac{u - 1}{u} d u

2 (- arctan (2) + 2)

十進法表記

1.78570 \dots

解答ステップ

\int_{1}^{5} \frac{u - 1}{u} d u

u置換積分法を適用する

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

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{v ^{2}}{v ^{2} + 1} d v

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

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

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}{v ^{2} + 1} d v + \int_{0}^{2} 1 d v)

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

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

= 2 (- arctan (2) + 2)