What is the integral from 1 to 2 of (1/t-i)^2 ?

The integral from 1 to 2 of (1/t-i)^2 is -1/2-i2ln(2)

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

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

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

Solution steps

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

Expand

(\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

Apply the Sum Rule:

\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)

y^{^{'}} = a y - b y^{2}

\frac{d}{d x} (\frac{x ^{8}}{16})

x \to 1 + lim (\frac{x + 2}{x + 1})

\int_{1}^{3} \frac{1}{x ( x + 1 )} d x

\int_{1}^{x^{2}} sin^{2} (t) d t

What is the integral from 1 to 2 of (1/t-i)^2 ?
The integral from 1 to 2 of (1/t-i)^2 is -1/2-i2ln(2)