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

The integral of sqrt(x^2+2x-3) is

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

= \int 4 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

= 4 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

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

Expand

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

= 4 \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

= 4 (- \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) ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to - 2 lim (x^{2} + 3)

f (x) = - 2 cos (x), at x = \frac{π}{4}

\frac{d}{d x} (e^{x} + x e^{x} - 1)

y d x - (2 y^{4} + 2 x) d y = 0

x^{4} - 8 x^{2} + 2