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

Lösung

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

\frac{e ^{2} - 1}{2} + 1

Dezimale

4.19452 \dots

Schritte zur Lösung

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

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

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

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

= \frac{e ^{2} - 1}{2} + 1

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