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

The integral of 1/(sqrt(x^2-8x-9)) is ln(1/5 |sqrt(x^2-8x-9)+x-4|)+C

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

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

ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4) + C

Solution steps

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

Complete the square

x^{2} - 8 x - 9 : (x - 4)^{2} - 25

= \int \frac{1}{( x - 4 ) ^{2} - 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 25} d u

= \int \frac{1}{u ^{2} - 25} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arcsec (\frac{1}{5} (x - 4))) + sec (arcsec (\frac{1}{5} (x - 4)))

Simplify

ln tan (arcsec (\frac{1}{5} (x - 4))) + sec (arcsec (\frac{1}{5} (x - 4))) : ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4)

= ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4)

Add a constant to the solution

= ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4) + C

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What is the integral of 1/(sqrt(x^2-8x-9)) ?
The integral of 1/(sqrt(x^2-8x-9)) is ln(1/5 |sqrt(x^2-8x-9)+x-4|)+C