What is the integral of sqrt((x^2)/4-1) ?

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

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integral of sqrt((x^2)/4-1)

\int \frac{x ^{2}}{4} - 1 d x

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

Solution steps

\int \frac{x ^{2}}{4} - 1 d x

Apply Trigonometric Substitution

= \int 2 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

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

Rewrite using trig identities

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

Expand

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

= 2 \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

= 2 (- \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) ∣

= 2 (- 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{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

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