What is the integral of 1/(sqrt(x^2-2x-63)) ?

The integral of 1/(sqrt(x^2-2x-63)) is ln(1/8 |sqrt(x^2-2x-63)+x-1|)+C

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integral of 1/(sqrt(x^2-2x-63))

\int \frac{1}{x ^{2} - 2 x - 63} d x

ln (\frac{1}{8} x^{2} - 2 x - 63 + x - 1) + C

Solution steps

\int \frac{1}{x ^{2} - 2 x - 63} d x

Complete the square

x^{2} - 2 x - 63 : (x - 1)^{2} - 64

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 64} d u

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

Simplify

ln tan (arcsec (\frac{1}{8} (x - 1))) + sec (arcsec (\frac{1}{8} (x - 1))) : ln (\frac{1}{8} x^{2} - 2 x - 63 + x - 1)

= ln (\frac{1}{8} x^{2} - 2 x - 63 + x - 1)

Add a constant to the solution

= ln (\frac{1}{8} x^{2} - 2 x - 63 + x - 1) + C

\int_{\frac{8}{3}}^{4} 3 t - 8 d t

\frac{d}{d x} (ln (\frac{a x + 1}{a x - 1}))

f (x) = 3 x + \frac{3}{x}

\frac{d}{d x} (\frac{e ^{x}}{1 - x})

f (x) = (x - \frac{2}{x})

What is the integral of 1/(sqrt(x^2-2x-63)) ?
The integral of 1/(sqrt(x^2-2x-63)) is ln(1/8 |sqrt(x^2-2x-63)+x-1|)+C