What is the integral from 1 to 4 of (sqrt(x^2-1))/x ?

The integral from 1 to 4 of (sqrt(x^2-1))/x is -arctan(sqrt(15))+sqrt(15)

Popular >

integral from 1 to 4 of (sqrt(x^2-1))/x

\int_{1}^{4} \frac{x ^{2} - 1}{x} d x

- arctan (15) + 15

Decimal

2.55486 \dots

Solution steps

\int_{1}^{4} \frac{x ^{2} - 1}{x} d x

Apply u-substitution

: \int_{0}^{15} \frac{u ^{2}}{u ^{2} + 1} d u

= \int_{0}^{15} \frac{u ^{2}}{u ^{2} + 1} d u

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

= \int_{0}^{15} - \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}^{15} \frac{1}{u ^{2} + 1} d u + \int_{0}^{15} 1 d u

\int_{0}^{15} \frac{1}{u ^{2} + 1} d u = arctan (15)

\int_{0}^{15} 1 d u = 15

= - arctan (15) + 15

x \to 9 lim (\frac{5 x}{x - 9})

f (x) = ((x^{2} + 7) (5 - x))

\frac{\partial}{\partial x} (e^{x - y})

\frac{x - 4}{x}

f (x) = x^{3} + x - 3

What is the integral from 1 to 4 of (sqrt(x^2-1))/x ?
The integral from 1 to 4 of (sqrt(x^2-1))/x is -arctan(sqrt(15))+sqrt(15)