What is the integral of 1/(sqrt((x-1)^2-4)) ?

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

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integral of 1/(sqrt((x-1)^2-4))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 4} d u

= \int \frac{1}{u ^{2} - 4} 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}{2} (x - 1))) + sec (arcsec (\frac{1}{2} (x - 1)))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (6 x ln (x) + 5 x)

f (x) = 2 x^{4} - 4 x^{2} + 6, at x = - 2

\frac{\partial}{\partial z} (tan (1 + 2 x^{2} y^{3} z^{2}))

\int_{- π}^{π} e^{- in x} d x

f (x) = \frac{4}{x}

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