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

The integral from 7 to 8 of (1/(x^2)-1) is -55/56

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

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

- \frac{55}{56}

Decimal Notation

- 0.98214 \dots

Solution steps

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

\int_{7}^{8} \frac{1}{x ^{2}} d x = \frac{1}{56}

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

= \frac{1}{56} - 1

Simplify

\frac{1}{56} - 1 : - \frac{55}{56}

= - \frac{55}{56}

(\frac{ln ( x )}{x ^{2}})^{^{'}}

\frac{\partial}{\partial x} (cos^{7} (x^{9} y^{5}))

y = \frac{4 - x e ^{x}}{x + e ^{x}}

\int x^{2} \frac{1}{x} d x

\int_{4}^{7} \frac{1}{x ^{2} - 1} d x

What is the integral from 7 to 8 of (1/(x^2)-1) ?
The integral from 7 to 8 of (1/(x^2)-1) is -55/56