What is the integral of sqrt(4x^2-16) ?

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

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integral of sqrt(4x^2-16)

\int 4 x^{2} - 16 d x

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

Solution steps

\int 4 x^{2} - 16 d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 8 \cdot \int - sec (u) + sec^{3} (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 sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 8 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{2}{4} x)

= 8 (- ln tan (arcsec (\frac{2}{4} x)) + sec (arcsec (\frac{2}{4} x)) + \frac{1}{2} sec (arcsec (\frac{2}{4} x)) tan (arcsec (\frac{2}{4} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{4} x)) + sec (arcsec (\frac{2}{4} x)))

Simplify

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

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

Add a constant to the solution

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

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