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

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

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

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

9 (- ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + \frac{1}{18} (x + 2) x^{2} + 4 x - 5 + \frac{1}{2} ln (\frac{x ^{2} + 4 x - 5 + x + 2}{3})) + C

Solution steps

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

Complete the square

x^{2} + 4 x - 5 : (x + 2)^{2} - 9

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

= \int 9 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

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

Rewrite using trig identities

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

Expand

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

= 9 \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

= 9 (- \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) ∣

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

Substitute back

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

Simplify

9 (- ln tan (arcsec (\frac{1}{3} (x + 2))) + sec (arcsec (\frac{1}{3} (x + 2))) + \frac{1}{2} sec (arcsec (\frac{1}{3} (x + 2))) tan (arcsec (\frac{1}{3} (x + 2))) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} (x + 2))) + sec (arcsec (\frac{1}{3} (x + 2)))) : 9 (- ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + \frac{1}{18} (x + 2) x^{2} + 4 x - 5 + \frac{1}{2} ln (\frac{x ^{2} + 4 x - 5 + x + 2}{3}))

= 9 (- ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + \frac{1}{18} (x + 2) x^{2} + 4 x - 5 + \frac{1}{2} ln (\frac{x ^{2} + 4 x - 5 + x + 2}{3}))

Add a constant to the solution

= 9 (- ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + \frac{1}{18} (x + 2) x^{2} + 4 x - 5 + \frac{1}{2} ln (\frac{x ^{2} + 4 x - 5 + x + 2}{3})) + C

x \to 7 + lim (\frac{1}{x + 7})

x \to 0 lim (\frac{2 sin ( 3 x )}{5 x})

\frac{\partial}{\partial x} (9 x^{7} cos (y^{4}))

f (x) = ln (64 sin^{2} (x))

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{3 t}