What is the integral from 1 to 2 of (z+2/z)*(z-2/z) ?

The integral from 1 to 2 of (z+2/z)*(z-2/z) is 1/3

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integral from 1 to 2 of (z+2/z)*(z-2/z)

\int_{1}^{2} (z + \frac{2}{z}) \cdot (z - \frac{2}{z}) d z

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

\int_{1}^{2} (z + \frac{2}{z}) (z - \frac{2}{z}) d z

Expand

(z + \frac{2}{z}) (z - \frac{2}{z}) : z^{2} - \frac{4}{z ^{2}}

= \int_{1}^{2} z^{2} - \frac{4}{z ^{2}} d z

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} z^{2} d z - \int_{1}^{2} \frac{4}{z ^{2}} d z

\int_{1}^{2} z^{2} d z = \frac{7}{3}

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

= \frac{7}{3} - 2

Simplify

\frac{7}{3} - 2 : \frac{1}{3}

= \frac{1}{3}

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What is the integral from 1 to 2 of (z+2/z)*(z-2/z) ?
The integral from 1 to 2 of (z+2/z)*(z-2/z) is 1/3