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

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

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

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

- \frac{7}{8}

Decimal Notation

- 0.875

Solution steps

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

\int_{7}^{8} \frac{7}{x ^{2}} d x = \frac{1}{8}

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

= \frac{1}{8} - 1

Simplify

\frac{1}{8} - 1 : - \frac{7}{8}

= - \frac{7}{8}

\int (\frac{x ^{2} + 1}{x ^{2}}) d x

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

\frac{d}{d x} ((x^{2} + 2 x)^{5})

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