What is the integral of sqrt(3x^2-4x+1) ?

The integral of sqrt(3x^2-4x+1) is 1/(6sqrt(3))((3x-2)sqrt(9x^2-12x+3)-ln|3x-2+sqrt(9x^2-12x+3)|)+C

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integral of sqrt(3x^2-4x+1)

\int 3 x^{2} - 4 x + 1 d x

\frac{1}{6 3} ((3 x - 2) 9 x^{2} - 12 x + 3 - ln 3 x - 2 + 9 x^{2} - 12 x + 3) + C

Solution steps

\int 3 x^{2} - 4 x + 1 d x

Complete the square

3 x^{2} - 4 x + 1 : 3 (x - \frac{2}{3})^{2} - \frac{1}{3}

= \int 3 (x - \frac{2}{3})^{2} - \frac{1}{3} d x

Apply u-substitution

: \int \frac{9 u ^{2} - 1}{3} d u

= \int \frac{9 u ^{2} - 1}{3} d u

\frac{9 u ^{2} - 1}{3} = \frac{1}{3} 9 u^{2} - 1

= \int \frac{1}{3} 9 u^{2} - 1 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 9 u^{2} - 1 d u

Apply Trigonometric Substitution

: \int \frac{1}{3} tan^{2} (v) sec (v) d v

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Expand

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

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

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} \frac{1}{3} (- \int sec (v) d v + \int sec^{3} (v) d v)

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

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

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

Substitute back

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

Simplify

\frac{1}{3} \frac{1}{3} (- ln tan (arcsec (3 (x - \frac{2}{3}))) + sec (arcsec (3 (x - \frac{2}{3}))) + \frac{1}{2} sec (arcsec (3 (x - \frac{2}{3}))) tan (arcsec (3 (x - \frac{2}{3}))) + \frac{1}{2} ln tan (arcsec (3 (x - \frac{2}{3}))) + sec (arcsec (3 (x - \frac{2}{3})))) : \frac{1}{6 3} ((3 x - 2) 9 x^{2} - 12 x + 3 - ln 3 x - 2 + 9 x^{2} - 12 x + 3)

= \frac{1}{6 3} ((3 x - 2) 9 x^{2} - 12 x + 3 - ln 3 x - 2 + 9 x^{2} - 12 x + 3)

Add a constant to the solution

= \frac{1}{6 3} ((3 x - 2) 9 x^{2} - 12 x + 3 - ln 3 x - 2 + 9 x^{2} - 12 x + 3) + C

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What is the integral of sqrt(3x^2-4x+1) ?
The integral of sqrt(3x^2-4x+1) is 1/(6sqrt(3))((3x-2)sqrt(9x^2-12x+3)-ln|3x-2+sqrt(9x^2-12x+3)|)+C