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

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

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

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

4 (- \frac{π}{3} + 3)

Decimal

2.73941 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int_{0}^{3} \frac{u ^{2}}{u ^{2} + 1} d u

= 4 \cdot \int_{0}^{3} \frac{u ^{2}}{u ^{2} + 1} d u

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

= 4 \cdot \int_{0}^{3} - \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

= 4 (- \int_{0}^{3} \frac{1}{u ^{2} + 1} d u + \int_{0}^{3} 1 d u)

\int_{0}^{3} \frac{1}{u ^{2} + 1} d u = \frac{π}{3}

\int_{0}^{3} 1 d u = 3

= 4 (- \frac{π}{3} + 3)

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\frac{\partial}{\partial x} (z ln ∣ x ∣)

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