What is the integral of 1/(sqrt(9x^2-12x-5)) ?

The integral of 1/(sqrt(9x^2-12x-5)) is 1/3 ln|sqrt((x-2/3)^2-1)+x-2/3 |+C

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integral of 1/(sqrt(9x^2-12x-5))

\int \frac{1}{9 x ^{2} - 12 x - 5} d x

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

Solution steps

\int \frac{1}{9 x ^{2} - 12 x - 5} d x

Complete the square

9 x^{2} - 12 x - 5 : 9 (x - \frac{2}{3})^{2} - 9

= \int \frac{1}{9 ( x - \frac{2}{3} ) ^{2} - 9} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(sqrt(9x^2-12x-5)) ?
The integral of 1/(sqrt(9x^2-12x-5)) is 1/3 ln|sqrt((x-2/3)^2-1)+x-2/3 |+C