What is the integral of 6zsqrt(4+z^2) ?

The integral of 6zsqrt(4+z^2) is 2(z^2+4)^{3/2}+C

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integral of 6zsqrt(4+z^2)

\int 6 z 4 + z^{2} d z

2 (z^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int 6 z 4 + z^{2} d z

Take the constant out:

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

= 6 \cdot \int z 4 + z^{2} d z

Apply u-substitution

: \int u^{2} d u

= 6 \cdot \int u^{2} d u

Apply the Power Rule

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

= 6 \cdot \frac{u ^{3}}{3}

Substitute back

u = 4 + z^{2}

= 6 \cdot \frac{( 4 + z ^{2} ) ^{3}}{3}

Simplify

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

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

Add a constant to the solution

= 2 (z^{2} + 4)^{\frac{3}{2}} + C

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What is the integral of 6zsqrt(4+z^2) ?
The integral of 6zsqrt(4+z^2) is 2(z^2+4)^{3/2}+C