What is the integral of 9zsqrt(3z^2-7) ?

The integral of 9zsqrt(3z^2-7) is (3z^2-7)^{3/2}+C

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integral of 9zsqrt(3z^2-7)

\int 9 z 3 z^{2} - 7 d z

(3 z^{2} - 7)^{\frac{3}{2}} + C

Solution steps

\int 9 z 3 z^{2} - 7 d z

Take the constant out:

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

= 9 \cdot \int z 3 z^{2} - 7 d z

Apply u-substitution

: \int \frac{u}{6} d u

= 9 \cdot \int \frac{u}{6} d u

Take the constant out:

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

= 9 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

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

= 9 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 z^{2} - 7

= 9 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 z^{2} - 7)^{\frac{3}{2}}

Simplify

9 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 z^{2} - 7)^{\frac{3}{2}} : (3 z^{2} - 7)^{\frac{3}{2}}

= (3 z^{2} - 7)^{\frac{3}{2}}

Add a constant to the solution

= (3 z^{2} - 7)^{\frac{3}{2}} + C

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What is the integral of 9zsqrt(3z^2-7) ?
The integral of 9zsqrt(3z^2-7) is (3z^2-7)^{3/2}+C