What is the integral of 1/(zsqrt(z^2-1)) ?

The integral of 1/(zsqrt(z^2-1)) is arctan(sqrt(z^2-1))+C

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integral of 1/(zsqrt(z^2-1))

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

arctan (z^{2} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= arctan (u)

Substitute back

u = z^{2} - 1

= arctan (z^{2} - 1)

Add a constant to the solution

= arctan (z^{2} - 1) + C

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What is the integral of 1/(zsqrt(z^2-1)) ?
The integral of 1/(zsqrt(z^2-1)) is arctan(sqrt(z^2-1))+C