What is the integral from 5 to 6 of (7/(x^2)-1) ?

The integral from 5 to 6 of (7/(x^2)-1) is -23/30

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integral from 5 to 6 of (7/(x^2)-1)

\int_{5}^{6} (\frac{7}{x ^{2}} - 1) d x

- \frac{23}{30}

Decimal Notation

- 0.76666 \dots

Solution steps

\int_{5}^{6} \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_{5}^{6} \frac{7}{x ^{2}} d x - \int_{5}^{6} 1 d x

\int_{5}^{6} \frac{7}{x ^{2}} d x = \frac{7}{30}

\int_{5}^{6} 1 d x = 1

= \frac{7}{30} - 1

Simplify

\frac{7}{30} - 1 : - \frac{23}{30}

= - \frac{23}{30}

\int cos (x) y

- 3 x^{3} - 2 x^{2} - 1

x \to 2 lim (x^{4} + 12 x + 8)

\frac{d}{d x} (\frac{5}{3} x^{\frac{2}{3}})

What is the integral from 5 to 6 of (7/(x^2)-1) ?
The integral from 5 to 6 of (7/(x^2)-1) is -23/30