integral from 1 to e of x/(x+1)

Lösung

\int_{1}^{e} \frac{x}{x + 1} d x

e - 1 - ln (e + 1) + ln (2)

Dezimale

1.09816 \dots

Schritte zur Lösung

\int_{1}^{e} \frac{x}{x + 1} d x

Wende U-Substitution an

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

Multipliziere aus

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

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

\int_{2}^{e + 1} 1 d u = e - 1

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

= e - 1 - (ln (e + 1) - ln (2))

Vereinfache

= e - 1 - ln (e + 1) + ln (2)