What is the integral of (sqrt(x^2-64))/x ?

The integral of (sqrt(x^2-64))/x is -8arcsec(1/8 x)+sqrt(x^2-64)+C

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integral of (sqrt(x^2-64))/x

\int \frac{x ^{2} - 64}{x} d x

- 8 arcsec (\frac{1}{8} x) + x^{2} - 64 + C

Solution steps

\int \frac{x ^{2} - 64}{x} d x

Apply Trigonometric Substitution

: \int 8 tan^{2} (u) d u

= \int 8 tan^{2} (u) d u

Take the constant out:

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

= 8 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 8 \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

= 8 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 8 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{8} x)

= 8 (- arcsec (\frac{1}{8} x) + tan (arcsec (\frac{1}{8} x)))

Simplify

8 (- arcsec (\frac{1}{8} x) + tan (arcsec (\frac{1}{8} x))) : - 8 arcsec (\frac{1}{8} x) + x^{2} - 64

= - 8 arcsec (\frac{1}{8} x) + x^{2} - 64

Add a constant to the solution

= - 8 arcsec (\frac{1}{8} x) + x^{2} - 64 + C

\frac{1 + 2 x}{3 - 4 x}

\frac{\partial}{\partial x} (\frac{V ( I , x )}{I})

\int e^{3 x + 2 y}

\int tan^{2} (x) sin (x) d x

f (x) = 7 x^{4}

What is the integral of (sqrt(x^2-64))/x ?
The integral of (sqrt(x^2-64))/x is -8arcsec(1/8 x)+sqrt(x^2-64)+C