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

Lösung

\int_{1}^{4} (1 + \frac{1}{u} + \frac{1}{u ^{2}}) d u

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

Dezimale

5.13629 \dots

Schritte zur Lösung

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

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

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

\int_{1}^{4} \frac{1}{u ^{2}} d u = \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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