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

The integral from 1 to 8 of (sqrt(x^2-1))/x is -arctan(3sqrt(7))+3sqrt(7)

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

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

- arctan (37) + 37

Decimal Notation

6.49178 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{37} \frac{u ^{2}}{u ^{2} + 1} d u

= \int_{0}^{37} \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= \int_{0}^{37} - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{37} \frac{1}{u ^{2} + 1} d u + \int_{0}^{37} 1 d u

\int_{0}^{37} \frac{1}{u ^{2} + 1} d u = arctan (37)

\int_{0}^{37} 1 d u = 37

= - arctan (37) + 37

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What is the integral from 1 to 8 of (sqrt(x^2-1))/x ?
The integral from 1 to 8 of (sqrt(x^2-1))/x is -arctan(3sqrt(7))+3sqrt(7)