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

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

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

\int x (3 x - 2) d x

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

Solution steps

\int x (3 x - 2) d x

Complete the square

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

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

= \frac{1}{3} \cdot \int \frac{1}{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

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

Rewrite using trig identities

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

Expand

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

= \frac{1}{3} \frac{1}{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

= \frac{1}{3} \frac{1}{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) ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (2 x^{2})

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

y = 4 x^{3} - 2 x

x \to 5 lim (x (5))

\frac{d}{d x} (\frac{- 2}{x})

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