What is the integral of sqrt(3x^2-1) ?

The integral of sqrt(3x^2-1) is 1/2 xsqrt(3x^2-1)-1/(2sqrt(3))ln|sqrt(3)x+sqrt(3x^2-1)|+C

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integral of sqrt(3x^2-1)

\int 3 x^{2} - 1 d x

\frac{1}{2} x 3 x^{2} - 1 - \frac{1}{2 3} ln 3 x + 3 x^{2} - 1 + C

Solution steps

\int 3 x^{2} - 1 d x

Apply Trigonometric Substitution

= \int \frac{1}{3} 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

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

Rewrite using trig identities

= \frac{1}{3} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= \frac{1}{3} \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

= \frac{1}{3} (- \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) ∣

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

Substitute back

u = arcsec (3 x)

= \frac{1}{3} (- ln tan (arcsec (3 x)) + sec (arcsec (3 x)) + \frac{1}{2} sec (arcsec (3 x)) tan (arcsec (3 x)) + \frac{1}{2} ln tan (arcsec (3 x)) + sec (arcsec (3 x)))

Simplify

\frac{1}{3} (- ln tan (arcsec (3 x)) + sec (arcsec (3 x)) + \frac{1}{2} sec (arcsec (3 x)) tan (arcsec (3 x)) + \frac{1}{2} ln tan (arcsec (3 x)) + sec (arcsec (3 x))) : \frac{1}{2} x 3 x^{2} - 1 - \frac{1}{2 3} ln 3 x + 3 x^{2} - 1

= \frac{1}{2} x 3 x^{2} - 1 - \frac{1}{2 3} ln 3 x + 3 x^{2} - 1

Add a constant to the solution

= \frac{1}{2} x 3 x^{2} - 1 - \frac{1}{2 3} ln 3 x + 3 x^{2} - 1 + C

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What is the integral of sqrt(3x^2-1) ?
The integral of sqrt(3x^2-1) is 1/2 xsqrt(3x^2-1)-1/(2sqrt(3))ln|sqrt(3)x+sqrt(3x^2-1)|+C