What is the integral of (sqrt(4x^2-64))/x ?

The integral of (sqrt(4x^2-64))/x is 8(-arcsec(x/4)+1/4 sqrt(x^2-16))+C

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integral of (sqrt(4x^2-64))/x

\int \frac{4 x ^{2} - 64}{x} d x

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

Solution steps

\int \frac{4 x ^{2} - 64}{x} d x

Apply Trigonometric Substitution

= \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{2}{8} x)

= 8 (- arcsec (\frac{2}{8} x) + tan (arcsec (\frac{2}{8} x)))

Simplify

8 (- arcsec (\frac{2}{8} x) + tan (arcsec (\frac{2}{8} x))) : 8 (- arcsec (\frac{x}{4}) + \frac{1}{4} x^{2} - 16)

= 8 (- arcsec (\frac{x}{4}) + \frac{1}{4} x^{2} - 16)

Add a constant to the solution

= 8 (- arcsec (\frac{x}{4}) + \frac{1}{4} x^{2} - 16) + C

What is the integral of (sqrt(4x^2-64))/x ?
The integral of (sqrt(4x^2-64))/x is 8(-arcsec(x/4)+1/4 sqrt(x^2-16))+C