What is the integral from 4 to 8 of (sqrt(x^2-16))/x ?

The integral from 4 to 8 of (sqrt(x^2-16))/x is 4(-pi/3+sqrt(3))

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integral from 4 to 8 of (sqrt(x^2-16))/x

\int_{4}^{8} \frac{x ^{2} - 16}{x} d x

4 (- \frac{π}{3} + 3)

Decimal

2.73941 \dots

Solution steps

\int_{4}^{8} \frac{x ^{2} - 16}{x} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{3}} 4 tan^{2} (u) d u

= \int_{0}^{\frac{π}{3}} 4 tan^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{3}} tan^{2} (u) d u

Rewrite using trig identities

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

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

\int_{0}^{\frac{π}{3}} sec^{2} (u) d u = 3

= 4 (- \frac{π}{3} + 3)

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What is the integral from 4 to 8 of (sqrt(x^2-16))/x ?
The integral from 4 to 8 of (sqrt(x^2-16))/x is 4(-pi/3+sqrt(3))