What is the integral of sqrt((2x-1)^2-9) ?

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

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integral of sqrt((2x-1)^2-9)

\int (2 x - 1)^{2} - 9 d x

\frac{1}{4} (2 (2 x - 1) x^{2} - x - 2 - 9 ln (\frac{2 x + 2 x ^{2} - x - 2 - 1}{3})) + C

Solution steps

\int (2 x - 1)^{2} - 9 d x

Expand

(2 x - 1)^{2} - 9 : 4 x^{2} - 4 x - 8

= \int 4 x^{2} - 4 x - 8 d x

Complete the square

4 x^{2} - 4 x - 8 : 4 (x - \frac{1}{2})^{2} - 9

= \int 4 (x - \frac{1}{2})^{2} - 9 d x

Apply u-substitution

: \int 4 u^{2} - 9 d u

= \int 4 u^{2} - 9 d u

Apply Trigonometric Substitution

: \int \frac{9}{2} tan^{2} (v) sec (v) d v

= \int \frac{9}{2} 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{9}{2} \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \frac{9}{2} \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= \frac{9}{2} \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{9}{2} (- \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{9}{2} (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{9}{2} (- ln tan (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + sec (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + \frac{1}{2} sec (arcsec (\frac{2}{3} (x - \frac{1}{2}))) tan (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + sec (arcsec (\frac{2}{3} (x - \frac{1}{2}))))

Simplify

\frac{9}{2} (- ln tan (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + sec (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + \frac{1}{2} sec (arcsec (\frac{2}{3} (x - \frac{1}{2}))) tan (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (\frac{2}{3} (x - \frac{1}{2}))) + sec (arcsec (\frac{2}{3} (x - \frac{1}{2})))) : \frac{1}{4} (2 (2 x - 1) x^{2} - x - 2 - 9 ln (\frac{2 x + 2 x ^{2} - x - 2 - 1}{3}))

= \frac{1}{4} (2 (2 x - 1) x^{2} - x - 2 - 9 ln (\frac{2 x + 2 x ^{2} - x - 2 - 1}{3}))

Add a constant to the solution

= \frac{1}{4} (2 (2 x - 1) x^{2} - x - 2 - 9 ln (\frac{2 x + 2 x ^{2} - x - 2 - 1}{3})) + C

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