What is the integral of sqrt(9x^2+12x+8) ?

The integral of sqrt(9x^2+12x+8) is 1/6 ((3x+2)sqrt(9x^2+12x+8)+4ln((|3x+sqrt(9x^2+12x+8)+2|)/2))+C

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integral of sqrt(9x^2+12x+8)

\int 9 x^{2} + 12 x + 8 d x

\frac{1}{6} ((3 x + 2) 9 x^{2} + 12 x + 8 + 4 ln (\frac{3 x + 9 x ^{2} + 12 x + 8 + 2}{2})) + C

Solution steps

\int 9 x^{2} + 12 x + 8 d x

Complete the square

9 x^{2} + 12 x + 8 : 9 (x + \frac{2}{3})^{2} + 4

= \int 9 (x + \frac{2}{3})^{2} + 4 d x

Apply u-substitution

: \int 9 u^{2} + 4 d u

= \int 9 u^{2} + 4 d u

Apply Trigonometric Substitution

: \int \frac{4}{3} sec^{3} (v) d v

= \int \frac{4}{3} sec^{3} (v) d v

Take the constant out:

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

= \frac{4}{3} \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{4}{3} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

= \frac{4}{3} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{4}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{2} ( x + \frac{2}{3} ) ) ) sin ( arctan ( \frac{3}{2} ( x + \frac{2}{3} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} (x + \frac{2}{3}))) + sec (arctan (\frac{3}{2} (x + \frac{2}{3}))))

Simplify

\frac{4}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{2} ( x + \frac{2}{3} ) ) ) sin ( arctan ( \frac{3}{2} ( x + \frac{2}{3} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} (x + \frac{2}{3}))) + sec (arctan (\frac{3}{2} (x + \frac{2}{3})))) : \frac{1}{6} ((3 x + 2) 9 x^{2} + 12 x + 8 + 4 ln (\frac{3 x + 9 x ^{2} + 12 x + 8 + 2}{2}))

= \frac{1}{6} ((3 x + 2) 9 x^{2} + 12 x + 8 + 4 ln (\frac{3 x + 9 x ^{2} + 12 x + 8 + 2}{2}))

Add a constant to the solution

= \frac{1}{6} ((3 x + 2) 9 x^{2} + 12 x + 8 + 4 ln (\frac{3 x + 9 x ^{2} + 12 x + 8 + 2}{2})) + C

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What is the integral of sqrt(9x^2+12x+8) ?
The integral of sqrt(9x^2+12x+8) is 1/6 ((3x+2)sqrt(9x^2+12x+8)+4ln((|3x+sqrt(9x^2+12x+8)+2|)/2))+C