What is the integral from 3 to 6 of (sqrt(x^2-9))/x ?

The integral from 3 to 6 of (sqrt(x^2-9))/x is -pi+3sqrt(3)

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integral from 3 to 6 of (sqrt(x^2-9))/x

\int_{3}^{6} \frac{x ^{2} - 9}{x} d x

- π + 33

Decimal

2.05455 \dots

Solution steps

\int_{3}^{6} \frac{x ^{2} - 9}{x} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 3 \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

= 3 (- \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

= 3 (- \frac{π}{3} + 3)

Simplify

3 (- \frac{π}{3} + 3) : - π + 33

= - π + 33

x \to 0 lim (\frac{1}{x - 1})

\int (x - 1) (x^{2} - 2 x)^{2} d x

(x^{2} + 4)^{- 3}

\frac{\partial}{\partial y} (2 x^{3} + x^{2} y + 3 x y^{2})

y = x^{2} + 3

What is the integral from 3 to 6 of (sqrt(x^2-9))/x ?
The integral from 3 to 6 of (sqrt(x^2-9))/x is -pi+3sqrt(3)