What is the integral of zsqrt(z^2-5) ?

The integral of zsqrt(z^2-5) is 1/3 (z^2-5)^{3/2}+C

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integral of zsqrt(z^2-5)

\int z z^{2} - 5 d z

\frac{1}{3} (z^{2} - 5)^{\frac{3}{2}} + C

Solution steps

\int z z^{2} - 5 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} - 5

= \frac{( z ^{2} - 5 ) ^{3}}{3}

Simplify

\frac{( z ^{2} - 5 ) ^{3}}{3} : \frac{1}{3} (z^{2} - 5)^{\frac{3}{2}}

= \frac{1}{3} (z^{2} - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (z^{2} - 5)^{\frac{3}{2}} + C

cos^{2} (5 t)

\int_{0}^{\infty} \frac{3}{x ^{2} + 4} d x

x y y^{^{'}} = 2 (2 x^{2} + y^{2})

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

0.0018425 (t + 7)^{4.5}

What is the integral of zsqrt(z^2-5) ?
The integral of zsqrt(z^2-5) is 1/3 (z^2-5)^{3/2}+C