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

Lösung

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

x - 5 (ln ∣ x + 5 ∣ - ln (5))

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

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

Wende die Summenregel an:

\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))

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