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

The integral from 1 to 2 of (x-2/x)^2 is 1/3

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integral from 1 to 2 of (x-2/x)^2

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Simplify

x - \frac{2}{x} : \frac{x ^{2} - 2}{x}

= \int_{1}^{2} (\frac{x ^{2} - 2}{x})^{2} d x

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int_{1}^{2} \frac{( x ^{2} - 2 ) ^{2}}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{( x ^{2} - 2 ) ^{2}}{x} - \int - 4 (x^{2} - 2) d x]_{1}^{2}

= [- \frac{( x ^{2} - 2 ) ^{2}}{x} - \int - 4 (x^{2} - 2) d x]_{1}^{2}

\int - 4 (x^{2} - 2) d x = - 4 (\frac{x ^{3}}{3} - 2 x)

= [- \frac{( x ^{2} - 2 ) ^{2}}{x} - (- 4 (\frac{x ^{3}}{3} - 2 x))]_{1}^{2}

Simplify

= [- \frac{( x ^{2} - 2 ) ^{2}}{x} + 4 (\frac{x ^{3}}{3} - 2 x)]_{1}^{2}

Compute the boundaries:

\frac{1}{3}

= \frac{1}{3}

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What is the integral from 1 to 2 of (x-2/x)^2 ?
The integral from 1 to 2 of (x-2/x)^2 is 1/3