What is the integral of (sqrt(x^2-256))/x ?

The integral of (sqrt(x^2-256))/x is -16arcsec(1/16 x)+sqrt(x^2-256)+C

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integral of (sqrt(x^2-256))/x

\int \frac{x ^{2} - 256}{x} d x

- 16 arcsec (\frac{1}{16} x) + x^{2} - 256 + C

Solution steps

\int \frac{x ^{2} - 256}{x} d x

Apply Trigonometric Substitution

: \int 16 tan^{2} (u) d u

= \int 16 tan^{2} (u) d u

Take the constant out:

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

= 16 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 16 \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

= 16 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 16 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{16} x)

= 16 (- arcsec (\frac{1}{16} x) + tan (arcsec (\frac{1}{16} x)))

Simplify

16 (- arcsec (\frac{1}{16} x) + tan (arcsec (\frac{1}{16} x))) : - 16 arcsec (\frac{1}{16} x) + x^{2} - 256

= - 16 arcsec (\frac{1}{16} x) + x^{2} - 256

Add a constant to the solution

= - 16 arcsec (\frac{1}{16} x) + x^{2} - 256 + C

x = - 3, x = 4, y = e^{2 x}, y = 2 e^{- x}

\frac{d}{d x} (x^{m} a^{2} - x^{2})

\frac{\partial}{\partial x} (x^{2} + 8 x y + 17 y^{2} - 2 y + 1)

\frac{\partial}{\partial x} (sin (2 t))

\int (- 8 x^{7} y^{- 5} + 6 x^{6} y^{- 4}) d y

What is the integral of (sqrt(x^2-256))/x ?
The integral of (sqrt(x^2-256))/x is -16arcsec(1/16 x)+sqrt(x^2-256)+C