What is the integral of 1/(sqrt(9x^2+12x+8)) ?

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

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integral of 1/(sqrt(9x^2+12x+8))

\int \frac{1}{9 x ^{2} + 12 x + 8} d x

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

Solution steps

\int \frac{1}{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 \frac{1}{9 ( x + \frac{2}{3} ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 4} d u

= \int \frac{1}{9 u ^{2} + 4} d u

Apply Trigonometric Substitution

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = sin (2 x), \frac{π}{2}, π

\frac{d}{d x} (\frac{6 x - x ^{2}}{tan ( x )})

\int e^{9 x} d x

\int x^{4} \cdot e^{x^{5}} d x

\frac{d}{d x} (3 e^{3 x} cos (2 x))

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