integral from 0 to 1 of x/(1+x)

Lösung

\int_{0}^{1} \frac{x}{1 + x} d x

1 - ln (2)

Dezimale

0.30685 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{x}{1 + x} d x

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

= 1 - ln (2)

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