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

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

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integral of 1/(sqrt(y^2-1))

\int \frac{1}{y ^{2} - 1} d y

ln y^{2} - 1 + y + C

Solution steps

\int \frac{1}{y ^{2} - 1} d y

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (y)

= ln ∣ tan (arcsec (y)) + sec (arcsec (y)) ∣

Simplify

ln ∣ tan (arcsec (y)) + sec (arcsec (y)) ∣ : ln y^{2} - 1 + y

= ln y^{2} - 1 + y

Add a constant to the solution

= ln y^{2} - 1 + y + C

n = 2 \sum \infty \frac{1}{n ( n + 2 )}

x y^{^{'}} + 2 y = 9 x

n = 1 \sum \infty \frac{( 4 ) ^{n - 1}}{5 ^{n}}

4 ln (7 + 2 x^{2})

f (x) = \frac{1}{x ^{6}}

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