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

The integral from 1 to 9 of (sqrt(x^2-1))/x is -arctan(4sqrt(5))+4sqrt(5)

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

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

- arctan (45) + 45

Decimal Notation

7.48481 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{45} \frac{u ^{2}}{u ^{2} + 1} d u

= \int_{0}^{45} \frac{u ^{2}}{u ^{2} + 1} d u

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

= \int_{0}^{45} - \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}^{45} \frac{1}{u ^{2} + 1} d u + \int_{0}^{45} 1 d u

\int_{0}^{45} \frac{1}{u ^{2} + 1} d u = arctan (45)

\int_{0}^{45} 1 d u = 45

= - arctan (45) + 45

f (x) = x, g (x) = - x, x = 1, x = 4

\frac{d}{d x} (4 sin (x) + 4 cos (x))

1 - 49 x^{2} arccos (7 x)

x \to 2 lim (7 (x) + 10)

What is the integral from 1 to 9 of (sqrt(x^2-1))/x ?
The integral from 1 to 9 of (sqrt(x^2-1))/x is -arctan(4sqrt(5))+4sqrt(5)