What is the integral of 1/(sqrt(9x^2+6x-8)) ?

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

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integral of 1/(sqrt(9x^2+6x-8))

\int \frac{1}{9 x ^{2} + 6 x - 8} d x

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

Solution steps

\int \frac{1}{9 x ^{2} + 6 x - 8} d x

Complete the square

9 x^{2} + 6 x - 8 : 9 (x + \frac{1}{3})^{2} - 9

= \int \frac{1}{9 ( x + \frac{1}{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{1}{3})) + sec (arcsec (x + \frac{1}{3}))

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty n \frac{1}{5 ^{n}}

\int sec^{2} (3 x t a) n^{5} 3 x d x

y ln (x) \frac{d y}{d x} = (\frac{( y + 1 )}{x})^{2}

(x^{2} - 7)^{2}

\frac{\partial}{\partial x} (ln (cos^{2} (x) + 1))

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