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

The integral of zsqrt(z^2-1) is 1/3 (z^2-1)^{3/2}+C

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integral of zsqrt(z^2-1)

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = z^{2} - 1

= \frac{( z ^{2} - 1 ) ^{3}}{3}

Simplify

\frac{( z ^{2} - 1 ) ^{3}}{3} : \frac{1}{3} (z^{2} - 1)^{\frac{3}{2}}

= \frac{1}{3} (z^{2} - 1)^{\frac{3}{2}}

Add a constant to the solution

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

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What is the integral of zsqrt(z^2-1) ?
The integral of zsqrt(z^2-1) is 1/3 (z^2-1)^{3/2}+C