What is the integral from 7 to 8 of (9/(x^2)-1) ?

The integral from 7 to 8 of (9/(x^2)-1) is -47/56

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integral from 7 to 8 of (9/(x^2)-1)

\int_{7}^{8} (\frac{9}{x ^{2}} - 1) d x

- \frac{47}{56}

Solution steps

\int_{7}^{8} \frac{9}{x ^{2}} - 1 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_{7}^{8} \frac{9}{x ^{2}} d x - \int_{7}^{8} 1 d x

\int_{7}^{8} \frac{9}{x ^{2}} d x = \frac{9}{56}

\int_{7}^{8} 1 d x = 1

= \frac{9}{56} - 1

Simplify

\frac{9}{56} - 1 : - \frac{47}{56}

= - \frac{47}{56}

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What is the integral from 7 to 8 of (9/(x^2)-1) ?
The integral from 7 to 8 of (9/(x^2)-1) is -47/56