integral from 1 to 2 of (1/t-i)^2

Lösung

\int_{1}^{2} (\frac{1}{t} - i)^{2} d t

- \frac{1}{2} - i 2 ln (2)

Schritte zur Lösung

\int_{1}^{2} (\frac{1}{t} - i)^{2} d t

Multipliziere aus

(\frac{1}{t} - i)^{2} : \frac{1 - t ^{2}}{t ^{2}} - i \frac{2}{t}

= \int_{1}^{2} \frac{1 - t ^{2}}{t ^{2}} - i \frac{2}{t} d t

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}^{2} \frac{1 - t ^{2}}{t ^{2}} d t - \int_{1}^{2} i \frac{2}{t} d t

\int_{1}^{2} \frac{1 - t ^{2}}{t ^{2}} d t = - \frac{1}{2}

\int_{1}^{2} i \frac{2}{t} d t = i 2 ln (2)

= - \frac{1}{2} - i 2 ln (2)