What is the integral of (4u-6)sqrt(u^2-3u) ?

The integral of (4u-6)sqrt(u^2-3u) is 4/3 (u(u-3))^{3/2}+C

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integral of (4u-6)sqrt(u^2-3u)

\int (4 u - 6) u^{2} - 3 u d u

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

Solution steps

\int (4 u - 6) u^{2} - 3 u d u

Apply u-substitution

: \int \frac{v v ^{2} - 36}{16} d v

= \int \frac{v v ^{2} - 36}{16} d v

Take the constant out:

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

= \frac{1}{16} \cdot \int v v^{2} - 36 d v

Apply u-substitution

: \int w^{2} d w

= \frac{1}{16} \cdot \int w^{2} d w

Apply the Power Rule

: \frac{w ^{3}}{3}

= \frac{1}{16} \cdot \frac{w ^{3}}{3}

Substitute back

= \frac{1}{16} \cdot \frac{( ( 4 u - 6 ) ^{2} - 36 ) ^{3}}{3}

Simplify

\frac{1}{16} \cdot \frac{( ( 4 u - 6 ) ^{2} - 36 ) ^{3}}{3} : \frac{4}{3} (u (u - 3))^{\frac{3}{2}}

= \frac{4}{3} (u (u - 3))^{\frac{3}{2}}

Add a constant to the solution

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

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What is the integral of (4u-6)sqrt(u^2-3u) ?
The integral of (4u-6)sqrt(u^2-3u) is 4/3 (u(u-3))^{3/2}+C