integral from 0 to x of t/(t+5)

解

\int_{0}^{x} \frac{t}{t + 5} d t

x - 5 (ln ∣ x + 5 ∣ - ln (5))

解答ステップ

\int_{0}^{x} \frac{t}{t + 5} d t

u置換積分法を適用する

= \int_{5}^{x + 5} \frac{u - 5}{u} d u

拡張

\frac{u - 5}{u} : 1 - \frac{5}{u}

= \int_{5}^{x + 5} 1 - \frac{5}{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_{5}^{x + 5} 1 d u - \int_{5}^{x + 5} \frac{5}{u} d u

\int_{5}^{x + 5} 1 d u = x

\int_{5}^{x + 5} \frac{5}{u} d u = 5 (ln ∣ x + 5 ∣ - ln (5))

= x - 5 (ln ∣ x + 5 ∣ - ln (5))