What is the integral of sqrt(x^2+2x-7) ?

The integral of sqrt(x^2+2x-7) is

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integral of sqrt(x^2+2x-7)

\int x^{2} + 2 x - 7 d x

8 (- ln x^{2} + 2 x - 7 + x + 1 + \frac{3}{2} ln (2) + \frac{1}{16} (x + 1) x^{2} + 2 x - 7 + \frac{1}{4} (2 ln x^{2} + 2 x - 7 + x + 1 - 3 ln (2))) + C

Solution steps

\int x^{2} + 2 x - 7 d x

Complete the square

x^{2} + 2 x - 7 : (x + 1)^{2} - 8

= \int (x + 1)^{2} - 8 d x

Apply u-substitution

= \int u^{2} - 8 d u

Apply Trigonometric Substitution

= \int 8 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

= 8 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 8 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= 8 \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

= 8 (- \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) ∣

= 8 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 8 (- ln tan (arcsec (\frac{1}{2 2} (x + 1))) + sec (arcsec (\frac{1}{2 2} (x + 1))) + \frac{1}{2} sec (arcsec (\frac{1}{2 2} (x + 1))) tan (arcsec (\frac{1}{2 2} (x + 1))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2 2} (x + 1))) + sec (arcsec (\frac{1}{2 2} (x + 1))))

Simplify

8 (- ln tan (arcsec (\frac{1}{2 2} (x + 1))) + sec (arcsec (\frac{1}{2 2} (x + 1))) + \frac{1}{2} sec (arcsec (\frac{1}{2 2} (x + 1))) tan (arcsec (\frac{1}{2 2} (x + 1))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2 2} (x + 1))) + sec (arcsec (\frac{1}{2 2} (x + 1)))) : 8 (- ln x^{2} + 2 x - 7 + x + 1 + \frac{3}{2} ln (2) + \frac{1}{16} (x + 1) x^{2} + 2 x - 7 + \frac{1}{4} (2 ln x^{2} + 2 x - 7 + x + 1 - 3 ln (2)))

= 8 (- ln x^{2} + 2 x - 7 + x + 1 + \frac{3}{2} ln (2) + \frac{1}{16} (x + 1) x^{2} + 2 x - 7 + \frac{1}{4} (2 ln x^{2} + 2 x - 7 + x + 1 - 3 ln (2)))

Add a constant to the solution

= 8 (- ln x^{2} + 2 x - 7 + x + 1 + \frac{3}{2} ln (2) + \frac{1}{16} (x + 1) x^{2} + 2 x - 7 + \frac{1}{4} (2 ln x^{2} + 2 x - 7 + x + 1 - 3 ln (2))) + C