What is the integral of (sqrt(x^2-16))/x ?

The integral of (sqrt(x^2-16))/x is -4arcsec(1/4 x)+sqrt(x^2-16)+C

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integral of (sqrt(x^2-16))/x

\int \frac{x ^{2} - 16}{x} d x

- 4 arcsec (\frac{1}{4} x) + x^{2} - 16 + C

Solution steps

\int \frac{x ^{2} - 16}{x} d x

Apply Trigonometric Substitution

: \int 4 tan^{2} (u) d u

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

Rewrite using trig identities

= 4 \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

= 4 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 4 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{4} x)

= 4 (- arcsec (\frac{1}{4} x) + tan (arcsec (\frac{1}{4} x)))

Simplify

4 (- arcsec (\frac{1}{4} x) + tan (arcsec (\frac{1}{4} x))) : - 4 arcsec (\frac{1}{4} x) + x^{2} - 16

= - 4 arcsec (\frac{1}{4} x) + x^{2} - 16

Add a constant to the solution

= - 4 arcsec (\frac{1}{4} x) + x^{2} - 16 + C

\int \frac{w}{1 - w} d w

x \to 1 lim (\frac{2}{( x - 1 ) ^{4}})

\int u^{- 1} d u

\frac{\partial}{\partial x} (3 x^{2} + 5 y^{7} + 9 x y)

\int_{2}^{4} (1 - (x - 3)^{2}) d x

What is the integral of (sqrt(x^2-16))/x ?
The integral of (sqrt(x^2-16))/x is -4arcsec(1/4 x)+sqrt(x^2-16)+C