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

The integral from 1/2 to 2 of (1-1/(x^2)) is 0

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

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

0

Solution steps

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

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

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

= \frac{3}{2} - \frac{3}{2}

Simplify

= 0

\frac{d y}{d x} = y + x^{2}

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What is the integral from 1/2 to 2 of (1-1/(x^2)) ?
The integral from 1/2 to 2 of (1-1/(x^2)) is 0