What is the integral of zsqrt(4z^2+8) ?

The integral of zsqrt(4z^2+8) is 2/3 (z^2+2)^{3/2}+C

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integral of zsqrt(4z^2+8)

\int z 4 z^{2} + 8 d z

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

Solution steps

\int z 4 z^{2} + 8 d z

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 z^{2} + 8

= \frac{1}{8} \cdot \frac{2}{3} (4 z^{2} + 8)^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (4 z^{2} + 8)^{\frac{3}{2}} : \frac{2}{3} (z^{2} + 2)^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{( x - 4 )}{y - 6}

y = 10 + 9 x - x^{2}, y = - 3 x + 46, [6, 15 + \frac{1}{3}]

t y^{^{'}} = t^{3} + 7 t^{3} y

\frac{d}{d x} (tan (x) + cot (x))

\int \frac{9 + x}{x} d x

What is the integral of zsqrt(4z^2+8) ?
The integral of zsqrt(4z^2+8) is 2/3 (z^2+2)^{3/2}+C