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

The integral from 1 to 2 of (x-1/x)^2 is 5/6

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

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

\frac{5}{6}

Decimal Notation

0.83333 \dots

Solution steps

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

Expand

(x - \frac{1}{x})^{2} : x^{2} - 2 + \frac{1}{x ^{2}}

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

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} x^{2} d x - \int_{1}^{2} 2 d x + \int_{1}^{2} \frac{1}{x ^{2}} d x

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

\int_{1}^{2} 2 d x = 2

\int_{1}^{2} \frac{1}{x ^{2}} d x = \frac{1}{2}

= \frac{7}{3} - 2 + \frac{1}{2}

Simplify

\frac{7}{3} - 2 + \frac{1}{2} : \frac{5}{6}

= \frac{5}{6}

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What is the integral from 1 to 2 of (x-1/x)^2 ?
The integral from 1 to 2 of (x-1/x)^2 is 5/6