What is the integral of sqrt(1+4x+x^2) ?

The integral of sqrt(1+4x+x^2) is

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integral of sqrt(1+4x+x^2)

\int 1 + 4 x + x^{2} d x

3 (- ln \frac{1}{3} (x^{2} + 4 x + 1) + \frac{1}{3} (x + 2) + \frac{1}{6} (x + 2) x^{2} + 4 x + 1 + \frac{1}{2} ln \frac{x ^{2} + 4 x + 1}{3} + \frac{1}{3} (x + 2)) + C

Solution steps

\int 1 + 4 x + x^{2} d x

Complete the square

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

= \int (x + 2)^{2} - 3 d x

Apply u-substitution

: \int u^{2} - 3 d u

= \int u^{2} - 3 d u

Apply Trigonometric Substitution

: \int 3 tan^{2} (v) sec (v) d v

= \int 3 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

= 3 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 3 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= 3 \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

= 3 (- \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) ∣

= 3 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 3 (- 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

3 (- 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)))) : 3 (- ln \frac{1}{3} (x^{2} + 4 x + 1) + \frac{1}{3} (x + 2) + \frac{1}{6} (x + 2) x^{2} + 4 x + 1 + \frac{1}{2} ln \frac{x ^{2} + 4 x + 1}{3} + \frac{1}{3} (x + 2))

= 3 (- ln \frac{1}{3} (x^{2} + 4 x + 1) + \frac{1}{3} (x + 2) + \frac{1}{6} (x + 2) x^{2} + 4 x + 1 + \frac{1}{2} ln \frac{x ^{2} + 4 x + 1}{3} + \frac{1}{3} (x + 2))

Add a constant to the solution

= 3 (- ln \frac{1}{3} (x^{2} + 4 x + 1) + \frac{1}{3} (x + 2) + \frac{1}{6} (x + 2) x^{2} + 4 x + 1 + \frac{1}{2} ln \frac{x ^{2} + 4 x + 1}{3} + \frac{1}{3} (x + 2)) + C

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