What is the integral of sqrt(x^2-5) ?

The integral of sqrt(x^2-5) is 5(1/10 xsqrt(x^2-5)-1/2 ln| 1/(sqrt(5))x+sqrt(1/5 (x^2-5))|)+C

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integral of sqrt(x^2-5)

\int x^{2} - 5 d x

5 (\frac{1}{10} x x^{2} - 5 - \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (x^{2} - 5)) + C

Solution steps

\int x^{2} - 5 d x

Apply Trigonometric Substitution

: \int 5 tan^{2} (u) sec (u) d u

= \int 5 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

= 5 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 5 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 5 \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

= 5 (- \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) ∣

= 5 (- 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}{5} x)

= 5 (- ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) + \frac{1}{2} sec (arcsec (\frac{1}{5} x)) tan (arcsec (\frac{1}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)))

Simplify

5 (- ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) + \frac{1}{2} sec (arcsec (\frac{1}{5} x)) tan (arcsec (\frac{1}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x))) : 5 (\frac{1}{10} x x^{2} - 5 - \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (x^{2} - 5))

= 5 (\frac{1}{10} x x^{2} - 5 - \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (x^{2} - 5))

Add a constant to the solution

= 5 (\frac{1}{10} x x^{2} - 5 - \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (x^{2} - 5)) + C

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What is the integral of sqrt(x^2-5) ?
The integral of sqrt(x^2-5) is 5(1/10 xsqrt(x^2-5)-1/2 ln| 1/(sqrt(5))x+sqrt(1/5 (x^2-5))|)+C