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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int 2 u^{2} - 1 d u

= \int 2 u^{2} - 1 d u

Take the constant out:

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

= 2 \cdot \int u^{2} - 1 d u

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

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

= 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

= 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) ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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