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

The integral from 1 to 6 of (sqrt(x^2-1))/x is -arctan(sqrt(35))+sqrt(35)

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

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

- arctan (35) + 35

Decimal Notation

4.51273 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{35} \frac{u ^{2}}{u ^{2} + 1} d u

= \int_{0}^{35} \frac{u ^{2}}{u ^{2} + 1} d u

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

= \int_{0}^{35} - \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}^{35} \frac{1}{u ^{2} + 1} d u + \int_{0}^{35} 1 d u

\int_{0}^{35} \frac{1}{u ^{2} + 1} d u = arctan (35)

\int_{0}^{35} 1 d u = 35

= - arctan (35) + 35

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1 - 2 t

What is the integral from 1 to 6 of (sqrt(x^2-1))/x ?
The integral from 1 to 6 of (sqrt(x^2-1))/x is -arctan(sqrt(35))+sqrt(35)