integral from 0 to t of y/(y+2)

解

\int_{0}^{t} \frac{y}{y + 2} d y

t - 2 (ln ∣ 2 + t ∣ - ln (2))

解答ステップ

\int_{0}^{t} \frac{y}{y + 2} d y

u置換積分法を適用する

= \int_{2}^{2 + t} \frac{u - 2}{u} d u

拡張

\frac{u - 2}{u} : 1 - \frac{2}{u}

= \int_{2}^{2 + t} 1 - \frac{2}{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_{2}^{2 + t} 1 d u - \int_{2}^{2 + t} \frac{2}{u} d u

\int_{2}^{2 + t} 1 d u = t

\int_{2}^{2 + t} \frac{2}{u} d u = 2 (ln ∣ 2 + t ∣ - ln (2))

= t - 2 (ln ∣ 2 + t ∣ - ln (2))