integral from 0 to 8 of (2x-4)/(x+1)

Lösung

\int_{0}^{8} \frac{2 x - 4}{x + 1} d x

16 - 12 ln (3)

Dezimale

2.81665 \dots

Schritte zur Lösung

\int_{0}^{8} \frac{2 x - 4}{x + 1} d x

Multipliziere aus

\frac{2 x - 4}{x + 1} : \frac{2 x}{x + 1} - \frac{4}{x + 1}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{8} \frac{2 x}{x + 1} d x - \int_{0}^{8} \frac{4}{x + 1} d x

\int_{0}^{8} \frac{2 x}{x + 1} d x = 2 (8 - 2 ln (3))

\int_{0}^{8} \frac{4}{x + 1} d x = 8 ln (3)

= 2 (8 - 2 ln (3)) - 8 ln (3)

Vereinfache

2 (8 - 2 ln (3)) - 8 ln (3) : 16 - 12 ln (3)

= 16 - 12 ln (3)

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