What is the integral of sqrt(4x^2-12x+5) ?

The integral of sqrt(4x^2-12x+5) is 1/4 (2x-3)sqrt(4x^2-12x+5)-ln(1/2 |2x-3+sqrt(4x^2-12x+5)|)+C

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integral of sqrt(4x^2-12x+5)

\int 4 x^{2} - 12 x + 5 d x

\frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 5 - ln (\frac{1}{2} 2 x - 3 + 4 x^{2} - 12 x + 5) + C

Solution steps

\int 4 x^{2} - 12 x + 5 d x

Factor

4 x^{2} - 12 x + 5 : (2 x - 1) (2 x - 5)

= \int (2 x - 1) (2 x - 5) d x

Apply u-substitution

: \int \frac{u ( u - 4 )}{2} d u

= \int \frac{u ( u - 4 )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u (u - 4) d u

Complete the square

u (u - 4) : (u - 2)^{2} - 4

= \frac{1}{2} \cdot \int (u - 2)^{2} - 4 d u

Apply u-substitution

: \int v^{2} - 4 d v

= \frac{1}{2} \cdot \int v^{2} - 4 d v

Apply Trigonometric Substitution

: \int 4 tan^{2} (w) sec (w) d w

= \frac{1}{2} \cdot \int 4 tan^{2} (w) sec (w) d w

Take the constant out:

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

= \frac{1}{2} \cdot 4 \cdot \int tan^{2} (w) sec (w) d w

Rewrite using trig identities

= \frac{1}{2} \cdot 4 \cdot \int (- 1 + sec^{2} (w)) sec (w) d w

Expand

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

= \frac{1}{2} \cdot 4 \cdot \int - sec (w) + sec^{3} (w) d w

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}{2} \cdot 4 (- \int sec (w) d w + \int sec^{3} (w) d w)

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

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

= \frac{1}{2} \cdot 4 (- ln ∣ tan (w) + sec (w) ∣ + \frac{1}{2} sec (w) tan (w) + \frac{1}{2} ln ∣ tan (w) + sec (w) ∣)

Substitute back

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

Simplify

\frac{1}{2} \cdot 4 (- ln tan (arcsec (\frac{1}{2} (2 x - 1 - 2))) + sec (arcsec (\frac{1}{2} (2 x - 1 - 2))) + \frac{1}{2} sec (arcsec (\frac{1}{2} (2 x - 1 - 2))) tan (arcsec (\frac{1}{2} (2 x - 1 - 2))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} (2 x - 1 - 2))) + sec (arcsec (\frac{1}{2} (2 x - 1 - 2)))) : \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 5 - ln (\frac{1}{2} 2 x - 3 + 4 x^{2} - 12 x + 5)

= \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 5 - ln (\frac{1}{2} 2 x - 3 + 4 x^{2} - 12 x + 5)

Add a constant to the solution

= \frac{1}{4} (2 x - 3) 4 x^{2} - 12 x + 5 - ln (\frac{1}{2} 2 x - 3 + 4 x^{2} - 12 x + 5) + C

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What is the integral of sqrt(4x^2-12x+5) ?
The integral of sqrt(4x^2-12x+5) is 1/4 (2x-3)sqrt(4x^2-12x+5)-ln(1/2 |2x-3+sqrt(4x^2-12x+5)|)+C