What is the integral of (sqrt(x^2-9))/x ?

The integral of (sqrt(x^2-9))/x is -3arcsec(1/3 x)+sqrt(x^2-9)+C

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integral of (sqrt(x^2-9))/x

\int \frac{x ^{2} - 9}{x} d x

- 3 arcsec (\frac{1}{3} x) + x^{2} - 9 + C

Solution steps

\int \frac{x ^{2} - 9}{x} d x

Apply Trigonometric Substitution

: \int 3 tan^{2} (u) d u

= \int 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 tan^{2} (u) d u

Rewrite using trig identities

= 3 \cdot \int - 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 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= 3 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{3} x)

= 3 (- arcsec (\frac{1}{3} x) + tan (arcsec (\frac{1}{3} x)))

Simplify

3 (- arcsec (\frac{1}{3} x) + tan (arcsec (\frac{1}{3} x))) : - 3 arcsec (\frac{1}{3} x) + x^{2} - 9

= - 3 arcsec (\frac{1}{3} x) + x^{2} - 9

Add a constant to the solution

= - 3 arcsec (\frac{1}{3} x) + x^{2} - 9 + C

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

f (x) = 9 t^{2} + t^{3}

\int 4 sin^{2} (x) cos (x) d x

\frac{d}{d x} ((ln (x))^{\frac{1}{x}})

\int_{0}^{1} (x - \frac{1}{2})^{2} d x

What is the integral of (sqrt(x^2-9))/x ?
The integral of (sqrt(x^2-9))/x is -3arcsec(1/3 x)+sqrt(x^2-9)+C