What is the integral of sqrt(3x^2+4x+1) ?

The integral of sqrt(3x^2+4x+1) is 1/(6sqrt(3))((3x+2)sqrt(9x^2+12x+3)-ln|3x+2+sqrt(9x^2+12x+3)|)+C

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integral of sqrt(3x^2+4x+1)

\int 3 x^{2} + 4 x + 1 d x

\frac{1}{6 3} ((3 x + 2) 9 x^{2} + 12 x + 3 - ln 3 x + 2 + 9 x^{2} + 12 x + 3) + C

Solution steps

\int 3 x^{2} + 4 x + 1 d x

Factor

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

= \int (3 x + 1) (x + 1) d x

Apply u-substitution

: \int u (3 u - 2) d u

= \int u (3 u - 2) d u

Complete the square

u (3 u - 2) : 3 (u - \frac{1}{3})^{2} - \frac{1}{3}

= \int 3 (u - \frac{1}{3})^{2} - \frac{1}{3} d u

Apply u-substitution

: \int \frac{9 v ^{2} - 1}{3} d v

= \int \frac{9 v ^{2} - 1}{3} d v

\frac{9 v ^{2} - 1}{3} = \frac{1}{3} 9 v^{2} - 1

= \int \frac{1}{3} 9 v^{2} - 1 d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 9 v^{2} - 1 d v

Apply Trigonometric Substitution

: \int \frac{1}{3} tan^{2} (w) sec (w) d w

= \frac{1}{3} \cdot \int \frac{1}{3} tan^{2} (w) sec (w) d w

Take the constant out:

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

= \frac{1}{3} \frac{1}{3} \cdot \int tan^{2} (w) sec (w) d w

Rewrite using trig identities

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

Expand

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

= \frac{1}{3} \frac{1}{3} \cdot \int - sec (w) + sec^{3} (w) d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} \frac{1}{3} (- \int sec (w) d w + \int sec^{3} (w) d w)

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

\int sec^{3} (w) d w = \frac{1}{2} sec (w) tan (w) + \frac{1}{2} ln ∣ tan (w) + sec (w) ∣

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

Substitute back

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

Simplify

\frac{1}{3} \frac{1}{3} (- ln tan (arcsec (3 (x + 1 - \frac{1}{3}))) + sec (arcsec (3 (x + 1 - \frac{1}{3}))) + \frac{1}{2} sec (arcsec (3 (x + 1 - \frac{1}{3}))) tan (arcsec (3 (x + 1 - \frac{1}{3}))) + \frac{1}{2} ln tan (arcsec (3 (x + 1 - \frac{1}{3}))) + sec (arcsec (3 (x + 1 - \frac{1}{3})))) : \frac{1}{6 3} ((3 x + 2) 9 x^{2} + 12 x + 3 - ln 3 x + 2 + 9 x^{2} + 12 x + 3)

= \frac{1}{6 3} ((3 x + 2) 9 x^{2} + 12 x + 3 - ln 3 x + 2 + 9 x^{2} + 12 x + 3)

Add a constant to the solution

= \frac{1}{6 3} ((3 x + 2) 9 x^{2} + 12 x + 3 - ln 3 x + 2 + 9 x^{2} + 12 x + 3) + C

3 \frac{d y}{d x} + 15 y = 5

x \to 0 lim (1 + e^{x})

\frac{( s + 3 )}{s ^{2} + 3 s + 2}

y^{2} - 2 x = 2, x - y = 3

\int \frac{2}{x - 1} d x

What is the integral of sqrt(3x^2+4x+1) ?
The integral of sqrt(3x^2+4x+1) is 1/(6sqrt(3))((3x+2)sqrt(9x^2+12x+3)-ln|3x+2+sqrt(9x^2+12x+3)|)+C