What is the integral of 1/(sqrt(x^2-6x-16)) ?

The integral of 1/(sqrt(x^2-6x-16)) is ln(1/5 |sqrt(x^2-6x-16)+x-3|)+C

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integral of 1/(sqrt(x^2-6x-16))

\int \frac{1}{x ^{2} - 6 x - 16} d x

ln (\frac{1}{5} x^{2} - 6 x - 16 + x - 3) + C

Solution steps

\int \frac{1}{x ^{2} - 6 x - 16} d x

Complete the square

x^{2} - 6 x - 16 : (x - 3)^{2} - 25

= \int \frac{1}{( x - 3 ) ^{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 - 3))) + sec (arcsec (\frac{1}{5} (x - 3)))

Simplify

ln tan (arcsec (\frac{1}{5} (x - 3))) + sec (arcsec (\frac{1}{5} (x - 3))) : ln (\frac{1}{5} x^{2} - 6 x - 16 + x - 3)

= ln (\frac{1}{5} x^{2} - 6 x - 16 + x - 3)

Add a constant to the solution

= ln (\frac{1}{5} x^{2} - 6 x - 16 + x - 3) + C

n = 1 \sum \infty e^{- 6 n}

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

sin (4 x) cos (2 x)

\int \frac{1.2 x}{x ^{2} x ^{4} - 9} d x

(\frac{sin ( x )}{1 + x})^{^{'}}

What is the integral of 1/(sqrt(x^2-6x-16)) ?
The integral of 1/(sqrt(x^2-6x-16)) is ln(1/5 |sqrt(x^2-6x-16)+x-3|)+C