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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

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

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

Add a constant to the solution

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

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