What is the integral of 1/(sqrt(x^2-6x-7)) ?

The integral of 1/(sqrt(x^2-6x-7)) is ln(1/4 |sqrt(x^2-6x-7)+x-3|)+C

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integral of 1/(sqrt(x^2-6x-7))

\int \frac{1}{x ^{2} - 6 x - 7} d x

ln (\frac{1}{4} x^{2} - 6 x - 7 + x - 3) + C

Solution steps

\int \frac{1}{x ^{2} - 6 x - 7} d x

Complete the square

x^{2} - 6 x - 7 : (x - 3)^{2} - 16

= \int \frac{1}{( x - 3 ) ^{2} - 16} d x

Apply u-substitution

= \int \frac{1}{u ^{2} - 16} d u

Apply Trigonometric Substitution

= \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}{4} (x - 3))) + sec (arcsec (\frac{1}{4} (x - 3)))

Simplify

ln tan (arcsec (\frac{1}{4} (x - 3))) + sec (arcsec (\frac{1}{4} (x - 3))) : ln (\frac{1}{4} x^{2} - 6 x - 7 + x - 3)

= ln (\frac{1}{4} x^{2} - 6 x - 7 + x - 3)

Add a constant to the solution

= ln (\frac{1}{4} x^{2} - 6 x - 7 + x - 3) + C

What is the integral of 1/(sqrt(x^2-6x-7)) ?
The integral of 1/(sqrt(x^2-6x-7)) is ln(1/4 |sqrt(x^2-6x-7)+x-3|)+C