What is the integral of 1/(sqrt(x^2-16x)) ?

The integral of 1/(sqrt(x^2-16x)) is ln(1/8 |sqrt(x^2-16x)+x-8|)+C

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integral of 1/(sqrt(x^2-16x))

\int \frac{1}{x ^{2} - 16 x} d x

ln (\frac{1}{8} x^{2} - 16 x + x - 8) + C

Solution steps

\int \frac{1}{x ^{2} - 16 x} d x

Complete the square

x^{2} - 16 x : (x - 8)^{2} - 64

= \int \frac{1}{( x - 8 ) ^{2} - 64} d x

Apply u-substitution

= \int \frac{1}{u ^{2} - 64} 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}{8} (x - 8))) + sec (arcsec (\frac{1}{8} (x - 8)))

Simplify

ln tan (arcsec (\frac{1}{8} (x - 8))) + sec (arcsec (\frac{1}{8} (x - 8))) : ln (\frac{1}{8} x^{2} - 16 x + x - 8)

= ln (\frac{1}{8} x^{2} - 16 x + x - 8)

Add a constant to the solution

= ln (\frac{1}{8} x^{2} - 16 x + x - 8) + C

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What is the integral of 1/(sqrt(x^2-16x)) ?
The integral of 1/(sqrt(x^2-16x)) is ln(1/8 |sqrt(x^2-16x)+x-8|)+C