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

The integral from 1 to 7 of (sqrt(x^2-1))/x is -arctan(4sqrt(3))+4sqrt(3)

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

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

- arctan (43) + 43

Decimal Notation

5.50075 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{43} \frac{u ^{2}}{u ^{2} + 1} d u

= \int_{0}^{43} \frac{u ^{2}}{u ^{2} + 1} d u

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

= \int_{0}^{43} - \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}^{43} \frac{1}{u ^{2} + 1} d u + \int_{0}^{43} 1 d u

\int_{0}^{43} \frac{1}{u ^{2} + 1} d u = arctan (43)

\int_{0}^{43} 1 d u = 43

= - arctan (43) + 43

\frac{d}{d t} (t - 1)

f (x) = 645120

\int_{- \infty}^{0} z e^{4 z} d z

f (x) = (x - 5)^{5}, at x = 6

x + \frac{1}{x ^{2}}

What is the integral from 1 to 7 of (sqrt(x^2-1))/x ?
The integral from 1 to 7 of (sqrt(x^2-1))/x is -arctan(4sqrt(3))+4sqrt(3)