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}

\int_{0}^{1} 3 x + 5 d x

\int_{0}^{t} (t - w) e^{5 w} d w

\int_{0}^{4} e^{x} x^{2} d x

\int_{0}^{2} \frac{4}{5 x + 1} d x

\int_{- 3}^{1} e^{u + 3} d u