integral from 0 to 2 of x/(1+x)

Lösung

\int_{0}^{2} \frac{x}{1 + x} d x

2 - ln (3)

Dezimale

0.90138 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{x}{1 + x} d x

Wende U-Substitution an

= \int_{1}^{3} \frac{u - 1}{u} d u

Multipliziere aus

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

= \int_{1}^{3} 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}^{3} 1 d u - \int_{1}^{3} \frac{1}{u} d u

\int_{1}^{3} 1 d u = 2

\int_{1}^{3} \frac{1}{u} d u = ln (3)

= 2 - ln (3)