integral from 1 to 4 of 1+1/x+1/(x^2)

Lösung

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

2 ln (2) + \frac{15}{4}

Dezimale

5.13629 \dots

Schritte zur Lösung

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

\int_{1}^{4} 1 d x = 3

\int_{1}^{4} \frac{1}{x} d x = 2 ln (2)

\int_{1}^{4} \frac{1}{x ^{2}} d x = \frac{3}{4}

= 3 + 2 ln (2) + \frac{3}{4}

Ziehe Brüche zusammen

3 + \frac{3}{4} : \frac{15}{4}

= 2 ln (2) + \frac{15}{4}

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